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METALLURGICAL  CALCULATIONS 


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METALLURGICAL  CALCULATIONS 


BY 

JOSEPH  W.  RICHARDS,  A.C.,  Ph.D. 

Professor  of  Metallurgy  in  Lehigh  University, 

Secretary  (and  Past-President)  of  the  American  Electrochemical  Society, 
V ice-President,  American  Institute  of  Mining  Engineers,  Mem- 
ber of  the  U.  S.  Naval  Consulting  Board,  Author  of 
t( Aluminium,  Its  Metallurgy,  etc.1' 


ONE  VOLUME  EDITION 


PART  I 

Introduction,  Chemical  and  Thermal  Principles 

Problems  in  Combustion,  and 
Radiation  and  Conduction  of  Heat 


PART  II 

Applications  to  the  Metallurgy  of  Iron  and  Steel 


PART  III 

Applications  to  Other  Metals  (Non-ferrous  Metals) 

THIRD  IMPRESSION 

McGRAW-HILL  BOOK  COMPANY,  INC, 

NEW   YORK:    239  WEST  39TH  STREET 

LONDON:    6  &  8  BOUVERIE  ST.,  E.  C.  4 

1918 


COPYRIGHT,  1910,  1915,  1917,  19x8,  BY  THE 
McGRAw-HiLL  BOOK  COMPANY,  INC. 


COPYRIGHT,  1906,  1907,  1908,  BY  THE 
McGRAw  PUBLISHING  COMPANY. 


THK  MAPLE  PRESS  YORK  PA 


PREFACE 

While  issuing  Parts  I,  II  and  III  separately,  for  the  use  of 
students,  it  is  thought  that  most  persons,  particularly  practical 
metallurgists,  will  prefer  to  have  the  whole  work  in  one  volume. 

The  writer  has  endeavored  to  correct  all  the  mistakes  which 
crept  into  former  editions,  and  to  add  such  new  and  useful  phys- 
ical and  chemical  data  as  have  appeared  to  date.  Some  new 
features  are  the  tables  of  thermochemical  constants  (which  will 
assist  in  predicting  undetermined  heats  of  formation),  the  esti- 
mation of  many  latent  heats  of  fusion  and  vaporization  of  the 
elements  (which  data  can  be  used  pro  tern,  as  first  approxima- 
tions), and  the  evaluation  of  vapor  tension  formulas  for  the  ele- 
ments in  both  liquid  and  solid  states  (likewise  to  be  used  as 
first  approximations).  My  thanks  are  due  to  Mr.  Leonard  Buck 
and  Mr.  Y.  Takikawa,  students  in  metallurgy,  for  considerable 
assistance  given  in  calculating  and  completing  the  new  tabular 
data. 

The  author  is  grateful  to  many  friends  whose  kindly  criticisms 
have  pointed  out  mistakes  or  suggested  improvements.  He  ap- 
preciates the  careful  translations  which  have  been  made  into 
Italian  by  Sr.  Remo  Catani,  into  Russian  by  Engineer  Koshkin, 
into  German  by  Prof.  Neuman  and  Engineer  Brodal,  and  regrets 
that  the  breaking  out  of  the  great  war  has  interrupted  transla- 
tions already  begun  into  French  and  Spanish.  These  inter- 
national recognitions  are  simply  proofs  that  the  day  of  Quanti- 
tative Metallurgy  is  dawning  all  over  the  world,  and  that  the 
application  of  Metallurgical  Calculations  is  the  path  to  Metallurgical 
Efficiency. 

JOSEPH  W.  RICHARDS. 
LEHIGH  UNIVERSITY, 
January  15,  1918. 


CONTENTS. 

PAGE 

INTRODUCTION. — Scope  of  the  Treatise iii-xi 

CHAPTER  I. — THE  CHEMICAL  EQUATION , .  1-12 

Atomic  Weights 1 

Relative  Weights 2 

Relative  Volumes  of  Gases 2 

Exact  Weights  and  Exact  Volumes 3 

Weights  and  Volumes  of  Gases 4 

Corrections  for  Temperature 5 

Corrections  for  Pressure 7 

Corrections  for  Temperature  and  Pressure 7 

Problem  1. — Combustion  of  Coal 8 

Problem  2. — Combustion  of  Natural  Gas 10 

Problem  3. — Oxidation  in  a  Bessemer  Converter..  11 

CHAPTER  II. — THE  APPLICATIONS  OP  THERMOCHEMISTRY.  13-40 

Thermochemical  Nomenclature ,  13 

Thermochemical  Data 18-40 

Oxides ;. 18 

Hydrates 19 

Sulfides 21 

Selenides,  Tellurides 22 

Arsenides,  Antimonides,  Phosphides,  Nitrides ...  23 

Hydrides,  Hydrocarbons 24 

Carbides 25 

Silicides,  Fluorides 26 

Chlorides 27 

Carbonates 28 

Bi-Carbonates 29 

Nitrates 30 

Silicates 31 

Sulphates ' 32 

Bi-sulphates,  Phosphates,  Arsenates,  Tungstates.  33 

Borates,  Molybdates,  Titanates 34 

Manganates,  Aluminates,  Cyanides 35 

Cyanates,  Metallo-cyanides 36 

vii 


Viii  CONTENTS. 

Amalgams,  Alloys 37 

Thermochemical  Constants  of  Bases  and  Acids.  38 

CHAPTER  III. — THE  USE  OP  THE  THERMOCHEMICAL  DATA  .  41-60 

Simple  Combinations — Complex  Combinations 41 

Double  Decompositions . . . .  43 

Calorific  Power  of  Fuels 46 

Problem  4. — Combustion  of  Natural  Gas 47 

Dulong's  Formula 48 

The  Theoretical  Temperature  of  Combustion 50 

Specific  Heats  of  Gases  Produced 51 

Combustion  with  Heated  Fuel  or  Air 52 

Problem  5. — Calorific  Intensity  of  Natural  Gas .  54 

The  Eldred  Process  of  Combustion 55 

Temperatures  in  the  "Thermit  Process" 58 

CHAPTER  IV. — THE  THERMOCHEMISTRY  OF  HIGH  TEMPERA- 
TURES  61-117 

Reduction  of  Zinc  Oxide  by  Carbon 63 

General  Remarks 68 

Reduction  of  Iron  Oxide  by  Hydrogen 69 

Specific  Heats  of  the  Elements 70 

Latent  Heats  of  Fusion  of  the  Elements 71 

Latent  Heats  of  Vaporization  of  the  Elements 73 

Thermophysics  of  the  Elements 74 

Tables  of  Thermophysics  of  the  Elements 75-103 

Efficiency  of  Furnaces 104 

Problem  6. — Efficiency  of  a  Rockwell  Furnace. .  106 
Problem  7. — Efficiency  of  an  Amalgam  Retort. .  107 
Problem  8. — Efficiency  of  a  Zinc  Distilling  Fur- 
nace   108 

Thermophysics  of  Alloys 109 

Tables  of  Thermophysics  of  Alloys 109-113 

Problem  9. — Efficiency  of  a  Steel  Melting  Fur- 
nace   114 

Problem  10. — Efficiency  of  a  Siemen's  Furnace. .  114 
Problem  11. — Efficiency  of  a  Foundry  Air-fur- 
nace   114 

Problem  12.— Efficiency  of  a  Foundry  Cupola. .  115 

CHAPTER  V. — THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS  118-144 

Oxides. .  118 


CONTENTS.  ix 

Problem  13. — Efficiency  of  Alumina  Melting  Fur- 
nace   127 

Problem  14. — Reduction  of  Iron  Oxides  by  CO 

gas. 128 

Problem  15. — Reduction  of  Iron  Oxides  by  Car- 
bon   129 

Chlorides 131 

Bromides,  Iodides 133 

Fluorides,  Sulfides 134 

Compound  Sulfides 135 

Arsenides,  Antimonides,  Selenides,  Tellurides 136 

Cyanides,  Hyposulfites,  Sulfates 136 

Carbonates,  Nitrates 137 

Borates,  Phosphates,  Chromates,  Arsenates 138 

Chlorates,  Aluminates,  Titanates,  etc 139 

Silicates 139 

Miscellaneous  Materials 142 

Slags 144 

CHAPTER  VI.— ARTIFICIAL  FURNACE  GAS 145-184 

Simple  Producers 146 

Problem  16. — Investigation    of    a    Simple    Pro- 
ducer   149 

Problem  17. — Drying  of  Moist  Producer  Gas. .  .  154 

Mixed  Gas  Producers 158 

Problem  18. — Investigation  of  a  Morgan  Pro- 
ducer   163 

Mond  Gas 169 

Problem  19. — Calculations  on  a  Mond  Producer .  169 

Reactions  on  Pre-heating  Mond  Gas 175 

Water  Gas 177 

Problem  20.— On  a  Dellwick-Fleischer  Plant. ...  179 

Problem  21. — Blast  Required  for  D-F.  Plant ...  182 

CHAPTER  VII. — CHIMNEY  DRAFT  AND  FORCED  DRAFT  . .   185-199 

Chimney  Draft 185 

Available  Head 191 

Problem  22. — Design  of  Puddling  Furnace  Chim- 
ney   192 

Problem  23. — Raising  Steam  by  Waste  Heat ...  194 


X  CONTENTS. 

CHAPTER  VIII. — CONDUCTION  AND  RADIATION  OF  HEAT.  186-200 

Principles  of  Heat  Conduction 200 

Table  of  Heat  Conductivities  of  Metals 203 

Principles  of  Heat  Transfer 206 

Problem  24. — Cooling  of  Hot  Air  in  a  Tube 208 

Table  of  Heat  Conductivities  of  Insulating  Ma- 
terials   211 

Radiation .'.  213 

Table  of  Radiating  Capacity  (Emissivity) 215 

APPENDIX 217-231 

Problem  25. — Combustion  of  Hydrocarbons 217 

Problem  26. — Calorific  Values  of  Coal,  Oil  and  Gas..  219 

Problem  27. — Calorific  Values  of  Various  Gases 219 

Problem  28. — Combustion  of  Coal  under  a  Boiler. .  .  220 

Problem  29. — Combustion  of  Anthracite  Coal 221 

Problem  30. — Air  Required  for  Powdered  Coal 221 

Problem  31. — Efficiency  Test  on  a  Boiler 222 

Problem  32. — Temperature  of  Powdered-coal  Flame.  223 
Problem  33. — Temperature  at  Tuyeres  in  Blast  Fur- 
nace   223 

Problem  34. — Efficiency  of  Powdered  Coal  Firing, . .  224 

Problem  35. — Calculation  of  Chimney  Draft 224 

Problem  36. — Calorimeter  Test  of  Temperature 224 

Problem  37. — Calorimeter  Test  for  Specific  Heat 225 

Problem  38. — Economy  of  a  Feed- water  Heater 225 

Problem  39. — Working  of  Gas-engine 226 

Problem  40. — Efficiency  of  a  Gas  Producer 226 

Problem  41. — Temperature  of  Producer-gas  Flame. .  227 

Problem  42. — Drying  Kiln  for  Peat 227 

Problem  43. — Power  from  Waste  Coke-oven  Gases. .  228 

Problem  44. — Efficiency  of  By-product  Coke  Ovens.  229 

Problem  45. — Decomposition  of  Steam  in  a  Producer .  229 

Problem  46. — Chimney  for  a  Gas  Furnace 230 

Problem  47. — Loss  of  Heat  through  Sides  of  Chimney  231 


CONTENTS  TO  PART  II. 

PAGE 

CHAPTER  I. — BALANCE  SHEET  OF  THE  BLAST  FURNACE.  .235-249 

Fuel 236 

Ore 239 

Flux... 243 

Blast 244 

Problem  51. — Balance  of  a  Swedish  Furnace.  .  .       245 

CHAPTER  II. — CALCULATION  OF  A  FURNACE  CHARGE  ..  250-267 

Flux  and  Slag 251 

Problem  52. — Calculation  of  Flux  needed 259 

Comparison  of  Values  of  Fuels 261 

Comparison  of  Values  of  Fluxes 263 

Comparison  of  Values  of  Ores 265 

Problem  53.— Cost  of  Pig  Iron  calculated 267 

CHAPTER  III. — UTILIZATION  OF  FUEL  IN  THE  BLAST  FUR- 
NACE  268-280 

Problem  54. — Generation  of  Heat  in  Blast  Fur- 
nace   268 

Problem  55.— Efficiency  of  Hot  Blast  Stoves 270 

Problem  56.— Power  from  Blast  Furnace  Gases.  272 

Gruners  Ideal  Working 274 

Minimum  Carbon  necessary  in  the  Furnace 277 

CHAPTER  IV. — HEAT  BALANCE  SHEET  OF  BLAST  FUR- 
NACE   281-304 

Heat  received  and  developed 281 

Combustion  of  Carbon. — Illustration 281 

Sensible  Heat  in  Hot  Blast.— Illustration 284 

Formation  of  Pig  Iron 285 

Formation  of  Slag 286 

xi 


xii  CONTENTS  TO  PART  II. 

PAGE 

Heat  absorbed  and  disbursed 287 

Heat  in  Waste  Gases 287 

Heat  in  Slag 288 

Heat  in  Pig  Iron 289 

Heat  conducted  away  and  radiated 289 

Heat  in  Cooling  Water 290 

Heat  for  drying  and  dehydrating  charges 290 

Heat  for  decomposing  carbonates 291 

Heat  for  reduction  of  iron  oxides 292 

Heat  for  reduction  of  other  oxides 292 

Decomposition  of  Moisture  of  Blast 293 

Problem  57. — Complete  Heat  Balance  Sheet 294 

CHAPTER  V. — RATIONALE  OF  HOT-BLAST  AND  DRY-BLAST  305-312 

Problem  58. — (for  practice)  Heat  Balance  Sheet.  305 

Hot  Blast. — Temperatures  obtained 308 

Dried  Blast. — Increased  Temperatures  obtained 309 

Illustration:  Temperatures  in  a  specific  case. ...  310 

Table  of  temperatures  under  various  conditions.  312 

CHAPTER  VI. — PRODUCTION,  HEATING  AND  DRYING  OP 

BLAST 313-332 

Production  of  blast.     Work  required 313 

Problem  59. — Work  of  a  blowing  engine 315 

Measurement  of  pressure  of  blast.     Indicator  cards . .  317 

Heating  of  the  blast 318 

Problem  60. — Efficiency  of  an  iron-pipe  stove. .  .  320 

Problem  61. — Efficiency  of  fire-brick  stoves 322 

Drying  air  blast 325 

Illustration:  Increased  efficiency  of  blowing  en- 
gines  t 326 

Illustration:  Refrigeration  required 326 

Problem  62. — Efficiency  of  drying  apparatus.  . .  327 
Illustrations:  Calculation  of  saturation  tempera- 
ture   331 

Problem  63. — Cooling  compressed  air  by  river 

water 331 

CHAPTER  VII. — THE  BESSEMER  PROCESS. 333-350 

Air  required 333 

Problem  64. — Maximum  and  minimum  blast. .  334 


CONTENTS  TO  PART  II.  xiii 

PAGE 

Air  received 336 

Problem  65. — Volume  efficiency  of  blowing  plant  336 

Blast  pressure 340 

Illustration:  Back  pressure  in  converter 341 

Problem  66. — Distribution  of  blast  pressure ....  343 

Flux  and  Slag 346 

Illustration:  Lime  needed  in  a  basic  blow 347 

Recarburization 348 

Problem  67. — Recarburization  balance  sheet 349 

CHAPTER  VIII. — THERMO-CHEMISTRY  OF  THE  BESSEMER 

PROCESS. 351-367 

Elements  consumed 351 

Heat  balance  sheet — items 353 

Heat  developed  by  oxidation 356 

Heat  of  formation  of  slag 357 

Illustration:  Heat  of  formation  of  basic  slag 358 

Heat  in  escaping  gases 360 

Heat  conducted  to  the  air 362 

Heat  radiated 363 

Problem  68. — Complete  heat  balance  sheet 363 

CHAPTER  IX. — THE  TEMPERATURE  INCREMENT  IN  THE 

BESSEMER  CONVERTER 368-380 

Combustion  of  Silicon 369 

Combustion  of  Manganese 371 

Combustion  of  Iron 372 

Combustion  of  Titanium 373 

Combustion  of  Aluminum,  Nickel,  Chromium 374 

Combustion  of  Carbon 375 

Combustion  of  Phosphorus 377 

Resume' 379 

CHAPTER  X.— THE  OPEN-HEARTH  FURNACE 381-395 

Gas  producers 381 

Flues  to  furnace;  regenerators 382 

Problem  69. — Sizes  of  regenerators 384 

Valves  and  ports 388 

Problem  70. — Areas  of  gas  and  air  ports 389 

Laboratory  of  furnace 391 


XIV  CONTENTS  TO  PART  II. 

PAGE 

Problem  71. — Efficiency  of  laboratory 393 

Chimney  flues  and  chimney,  miscellaneous 395 

CHAPTER  XI. — THERMAL  EFFICIENCY  OF  OPEN-HEARTH 

FURNACES 396-428 

Heat  in  warm  charges ;  in  gas  used ' 397 

Heat  in  air  used 398 

Heat  of  combustion;  of  oxidation  of  the  bath 399 

Heat  of  formation  of  slag;  heat  in  melted  steel 399 

Reduction  of  iron  ore;  decomposition  of  carbonates .  .  400 

Heat  in  slag 401 

Loss  by  imperfect  combustion ;  in  chimney  gases ....  402 

Loss  by  conduction  and  radiation 403 

Problem  72. — Heat  balance  sheet. . . 403 

Problem  73.— Thermal  efficiencies 411 

Problem  74. — New  Siemens'  furnace 418 

Problem  75. — The  Monell  process 424 

CHAPTER  XII. — THE  ELECTROMETALLURGY  OF  IRON  AND 

STEEL > -.429-451 

Electrothermal  reduction  of  iron  ores 429 

Problem  76. — Production  of  pig-iron  electrically .  430 

Problem  77. — Production  of  nickelif erous  pig .  .  .  435 

Production  of  Steel 440 

Problem  78. — The  induction  electric  furnace.  .  .  442 

Problem  79. — The  electric  "  pig  and  ore  "  process  443 
Problem  80. — Electrolytic      refining;      Burgess' 

process 446 

APPENDIX:  PROBLEMS  FOR  PRACTICE 452-465 

Problem  81. — Diameter  of  tuyeres  for  a  blast  fur- 
nace    452 

Problem  82. — Blast  furnace  slag  calculation 452 

Problem  83. — Carbon  burned  before  the  tuyeres 453 

Problem  84. — Calculation  of  blast  furnace  charge.  .  .  454 

Problem  85. — Slag  calculation  for  various  slags 454 

Problem  86. — Weight  of  flux  needed 455 

Problem  87. — Volume  of  blast,  and  cooling  of  same 

by  expansion 456 

Problem  88. — Output  of  a  blast-furnace 456 


CONTENTS  TO  PART  II.  XV 

PAGE 

Problem    89. — Efficiency  of  the  blowing  engines 456 

Problem    90. — Power  producible  from  blast-furnace 

gases ^ 457 

Problem    91. — Proportion  of  gases  required  by  hot- 
blast  stoves • 457 

Problem    92. — Complete  balance  sheet  of  a  blast  fur- 
nace        458 

Problem    93. — Working  of  a  foundry  cupola 460 

Problem    94. — The  Monell  process  in  an  open-hearth 

furnace 460 

Problem    95. — Volume  of  blast  required  by  a  Besse- 
mer converter 461 

Problem    96. — Volume  of  blast  received  by  a  Besse- 
mer converter 461 

Problem    97. — Loss  of  weight  of  a  Bessemer  charge.       462 
Problem    98. — Balance  sheet  of  a  complete  Bessemer 

blow 462 

Problem    99. — Lime  needed  in  a  basic  Bessemer  blow      464 
Problem  100. — Power  of  blowing  engines  required  for 

a  converter 465 

INDEX  . , 466 


CONTENTS  TO  PART  III. 

PAGE 

CHAPTER  I. — THE  METALLURGY  OF  COPPER 469-577 

Roasting  and  Smelting  of  Copper  Ores - 469 

The  Roasting  Operation 475 

Problems  101-104. 

Pyritic  Smelting 483 

Fundamental  Principles 485 

Theoretical  Temperature  at  the  Focus 486 

Use  of  Auxiliary  Coke 488 

Rate  of  Smelting 489 

Problems  105,  106. 

Smelting  of  Copper  ores 499 

Reverberatory  Smelting 501 

Problems  107,  108. 

"Bringing  Forward"  of  Copper  Matte 514 

The  Welsh  Process 514 

Problem  109. 

"Blister  Roasting"  or  Roasting-Smelting 523 

"  Bessemerizing"  Copper  Matte 525 

Theoretical  Temperature  Rise 527 

Problem  110. 

The  Electrometallurgy  of  Copper 534 

Electrolytic  Processes 536 

Treatment  of  Ores  by  Electrolysis 536 

Problem  111. 

Electrolytic  Treatment  of  Matte 539 

Use  of  Matte  as  Anode 539 

Problem  112. 

Use  of  Matte  as  Cathode 545 

Problem  113. 

Extraction  from  Solutions 548 

(1)  Use  of  Iron  Anodes 548 

xvii 


xviil  CONTENTS  TO  PART  III. 

PAGE 

Problems  114,  115. 
(2)  Use  of  Insoluble  Anodes 551 

Problems  116,  117,  118 

Electrolytic  Refining  of  Impure  Copper 558 

Energy  Absorbed  by  Electrolyte ,       559 

Problem  119. 
Energy  Lost  in  Contacts 562 

Problem  120. 
Energy  Lost  in  Conductors 567 

Problem  121. 
Electric  Smelting 572 

Problem  122. 

CHAPTER  II. — THE  METALLURGY  OF  LEAD 578-604 

The  Volatility  of  Lead 583 

Roasting  of  Lead  Ores 586 

Problem  123. 
Reduction  of  Roasted  Ore 591 

Problem  124. 
The  Electrometallurgy  of  Lead 598 

Problems  125,  126. 

CHAPTER  III. — THE  METALLURGY  OF  SILVER  AND  GOLD. 605-619 

Electrolytic  Refining  of  Silver  Bullion 605 

Problems  127,  128. 
The  Volatilization  of  Silver  and  Gold 614 

CHAPTER  IV. — THE  METALLURGY  OF  ZINC 620-657 

Roasting  of  Sphalerite 620 

Problems  129,  130. 

Reduction  of  Zinc  Oxide 627 

Thermochemical  Considerations 628 

Heating  up  the  Charge 628 

Problem  131. 

Distillation  of  the  Charge 632 

Problems  132,  133,  134. 

Electric  Smelting  of  Zinc  Ores 637 

Zinc  Vapor 643 


CONTENTS  TO  PART  III.  xix 

PAGE 

Problem  135. 

Condensation  of  Zinc  and  Mercury 649 

Metallic  Mist  or  Fume  Vapor 656 

CHAPTER  V. — METALLURGY  OF  ALUMINIUM 658-661 

Electrolytic  furnace  reduction  of  alumina 658 

INDEX.  .  663 


METALLURGICAL  CALCULATIONS. 

INTRODUCTION. 

The  making  of  calculations  respecting  the  quantitative  work- 
ing of  any  process,  furnace  or  piece  of  apparatus  used  in  metal- 
lurgical operations  is  of  the  greatest  importance  for  estimating 
the  real  efficiency  of  the  process,  for  determining  avenues  of 
waste  and  possible  lines  of  improvement,  and  for  obtaining  the 
best  possible  comprehension  of  the  real  principles  of  operation 
involved. 

The  possibility  of  making  such  calculations  respecting  any 
process,  furnace  or  apparatus  depends  on  skill  in  collecting  such 
accessary  data  as  can  be  obtained  by  observation  or  measure- 
ment, the  insight  or  intuition  to  see  the  further  use  which  can 
be  made  of  §aid  data  when  once  obtained,  and,  finally  on  the 
possession  of  a  working  knowledge  of  the  fundamental  chem- 
ical, physical  and  mechanical  principles  involved  in  the  calcu- 
lations. The  highest  desideratum,  all  in  all,  however,  is  a 
plain  analytical,  common  sense  mind,  capable  of  clear,  logical 
thinking  It  is  the  writer's  conviction  that  no  study  of  details, 
or  even  observation  of  plants  in  actual  operation,  can  supply 
the  insight  into  metallurgical  processes  and  principles,  such 
as  is  gained  by  these,  calculations,  in  addition  to  the  high  grade 
of  mental  training  involved. 

SCOPE  OF  THE  TREATISE 

Discussion  of  the  chemical  equation. 
Weights  and  volumes  of  gases. 

Correction  of  gas  volumes  for  temperature  and  pressure. 
Combustion  of  commercial  fuels. 
Heat  of  chemical  combination;  of  combustion. 
Theoretical  flame  temperatures: 
With  pure  oxygen. 

With  ordinary  air. 

xxi 


xxii  INTRODUCTION. 

With  diluted  air:  Farley's  system. 

With  hot  air  and  cold  gas. 

With  hot  air  and  hot  gas. 

Effect  of  excess  air. 
Calculation  of  furnace  efficiencies. 
Chimney  draft. 
Water  gas. 

Producer  gas:  Efficiency,  effect  of  drying. 
Mixed  gas: 

Use  of  steam  in  producers. 

Increased  efficiency. 

Maximum  steam  permissible. 
Transmission  of  heat  through  metals,  brick,  etc. 
Regenerative  gas  furnace: 

Proportioning  of  gas  and  air  regenerators. 

Efficiency  of  regenerators. 

Heat  balance  sheet. 

Theoretical  temperatures  under  different  conditions. 
Gas  engines: 

Calculation  of  temperature  in  cylinder. 

Efficiency;  balance  sheet. 
Cupolas:  Amount  of  blast  required. 

Efficiency  of  running. 
Blast  furnaces: 

Balance  sheet  of  materials. 

Calculation  of  blast  received. 

Efficiency  of  blowing  engines. 

Power  and  dimensions  of  blowing  engines. 

Carbon  consumed  at  tuyeres. 

Effect  of  atmospheric  changes. 

Effect  of  the  moisture  in  the  blast. 

Calculation  of  the  temperature. 

Effect  of  hot-blast. 

Heat  balance  sheet  of  the  furnace. 

Power  producible  from  the  waste  gases. 
Hot-blast  stoves:  Theory  of  iron-pipe  and  fire-brick  stoves. 

Efficiency. 
Bessemer  Converters: 

Blast  required  and  time  of  operation 

Balance  sheet  of  materials. 


INTRODUCTION.  xxiii 

Heating  efficiency  of  various  ingredients  of  bath. 

Heat  balance  sheet. 

Theoretical  rise  in  temperature. 

Conversion  of  copper  matte. 
Open-hearth  Furnaces: 

Pig  and  ore  process;  calculation  of  charge. 

Heat  evolved  or  absorbed  in  bath  reactions. 

Efficiency  of  furnaces;  of  furnaces  and  producers. 

Heat  balance  sheet. 
Pyritic  smelting. 
Electric  furnaces: 

Working  temperatures. 

Heat  balance  sheet. 

Efficiency, 
Electrolytic  furnaces: 

Absorption  of  heat  in  chemical  decompositions. 

Equilibrium  of  temperature  attained. 

Ampere  and  energy  efficiency. 
Electrolytic  refining: 

Calculation  of  plant  and  output 

Power  requirements;  temperature  of  baths. 

Ampere  and  energy  efficiency. 
Condensation  of  metallic  vapors: 

Principles  involved. 

Application  to  condensation  of  zinc  and  mercury. 


CHAPTER  I. 


THE  CHEMICAL  EQUATION. 

The  calculation  of  the  quantitative  side  of  many  metallur- 
gical processes  depends  upon  the  correct  understanding  of 
chemical  equations.  Every  chemical  equation  is  capable  of 
giving  three  most  important  sets  of  data  concerning  the  pro- 
cess which  it  represents;  its  shows  the  relative  weights  of  the 
reacting  substances,  their  relative  volumes,  when  in  the  gas- 
eous state,  and  the  surplus  or  deficit  of  energy  involved  in  the 
reaction,  when  the  heats  of  formation  of  the  substances  con- 
cerned are  known. 

ATOMIC  WEIGHTS. 

These  are  the  basis  of  all  quantitative  chemical  calculations. 
For  metallurgical  purposes  we  may  use  them  in  round  num- 
bers as: 


Hydrogen H       1 

Lithium Li  7 

Beryllium Be  9 

Boron B  11 

Carbon C  12 

Nitrogen .  .N  14 

Oxygen O  16 

Fluorine F  19 

Sodium Na  23 

Magnesium Mg  24 

Aluminium Al  27 

Silicon Si  28 

Phosphorus P  31 

Sulphur S  32 

Chlorine Cl  35.5 

Potassium ". K  39 

Calcium..  .  .Ca  40 


Arsenic As  75 

Selenium Se  79 

Bromine. Br  80 

Strontium Sr  87 

Zirconium Zr  90 

Columbium Cb  94 

Molybdenum Mo  96 

Palladium Pd  106 

Silver Ag  108 

Cadmium Cd  112 

Tin Sn  118 

Antimony Sb  120 

Tellurium Te  126 

Iodine I  127 

Barium Ba  137 

Tantalum Ta  183 

Tungsten W  184 


2  METALLURGICAL  CALCULATIONS. 

Titanium Ti     48         Tridium Ir  193 

Vanadium V     51         Platinum Pt   195 

Chromium Cr       52         Gold Au   197 

Manganese Mn       55         Mercury Hg     200 

Iron .  .  .Fe       56         Thallium Tl     204 

Nickel ..Ni  58.5         Lead Pb     207 

Cobalt Co       59         Bismuth Bi     208 

Copper Cu  63.6         Thorium Th     232 

Zinc Zn       65         Uranium. U     238 

RELATIVE  WEIGHTS. 

Writing  any  chemical  equation  between  elements  or  their 
compounds,  the  relative  weights  concerned  in  the  reaction  are 
obtained  directly  from  using  these  atomic  weights,  which  are, 
themselves,  of  course,  only  relative.  E.g.,  the  slagging  of  iron 
in  Bessemerizing  copper  matte: 

2FeS  +  3O2  +  2SiO2  =  2(FeO.Si02)  +  2S02 
176+   96+     120=        264        +128 

These  relative  weights  may  be  called  kilograms  or  tons, 
pounds,  ounces  or  grains;  whatever  units  of  weight  we  may 
be  working  in.  In  most  metallurgical  work  we  use  kilograms 
or  pounds  as  the  convenient  weight  units. 

RELATIVE  VOLUMES  OF  GASES. 

Where  gases  are  involved,  the  relative  number  of  molecules 
of  the  gaseous  substance  concerned  in  the  reaction  stands  for 
the  relative  volume  of  that  gas  concerned  in  the  reaction.  It  is 
usual  and  convenient  to  designate  these  relative  volumes  by 
Roman  numerals,  placed  above  the  formulae.  The  following 
are  some  examples: 

Complete  combustion  of  carbon: 

i  i 

C  +  O2  =  CO2 


Incomplete  combustion  of  carbon: 

i 
2C  +  O 

Combustion  of  marsh  gas: 


i  n 

2C  +  O2  =  2CO 


i          n  i  ii 

CH4  +  2O2  =  CO2  +  2H,0 


THE  CHEMICAL  EQUATION.  3 

Production  of  water  gas: 

i  i         i 

C  +  H2O  =  CO  +  H2 

In  each  case  above,  the  volume  of  a  solid  or  liquid  cannot  be 
stated,  but  the  relative  volumes  of  all  the  gases  taking  part  in  a 
reaction  are  derived  simply  from  the  number  of  molecules  of 
each  gas  concerned.  These  relative  volumes  may  be  called  so 
many  cubic  meters  or  liters,  or  cubic  feet,  or  whatever  measure 
is  wanted  or  being  used.  In  most  metallurgical  calculations  it 
is  convenient  to  use  cubic  meters  or  cubic  feet. 

EXACT  WEIGHTS  AND  EXACT  VOLUMES. 

If  we  specify  or  fix  the  weights  used,  as,  for  instance,  so 
many  kilograms  of  each  substance  as  the  numbers  representing 
the  relative  weights,  then  we  can,  by  using  one  constant  factor, 
convert  all  the  relative  volumes  into  the  real  or  absolute  vol- 
umes corresponding  to  the  weights  used.  If,  for  instance,  we 
take  the  equation  of  the  production  of  water  gas: 

i         i         i 
C  +  H20  =  CO  +  H2 

12+   18  =  28  +  2 

With  the  relative  weights  written  beneath  and  the  relative 
volumes  above,  then  if  we  fix  the  weights  as  kilograms,  the 
relative  volumes  can  be  converted  into  actual  volumes  in  cubic 
meters  by  multiplying  by  22.22.  A  cubic  meter  of  hydrogen 
gas  (under  standard  conditions)  weighs  0.09  kilogram,  and 
thence  2  kilograms  will  have  a  volume  of  2-7-0.09  =  22.22  cubic 
meters.  But  the  relative  volumes  show  that  the  CO  and 
H2O  gas  are  the  same  in  volume  as  the  hydrogen,  and  it,  there- 
fore, follows  that  each  Roman  I  stands  for  22.22  cubic  meters 
of  gas,  if  the  weights  underneath  are  called  kilograms.  The 
consideration  of  these  relations  is  very  advantageous,  because, 
by  means  of  this  factor  (22.22)  we  pass  at  once  from  the 
weight  of  a  gas  to  its  volume;  each  molecule  or  molecular 
weight  of  a  gas,  in  kilograms  (or,  briefly,  each  kilogram-mole- 
cule), represents  22.22  cubic  meters  of  that  gas. 

The  conversion  from  weight  to  volume  is  quite  as  simple 
using  English  measures;  and,  by  a  strange  coincidence,  the 
same  factor  can  be  used  as  in  the  metric  system.  The  coin- 


4  METALLURGICAL  CALCULATIONS. 

cidence  alluded  to  is  the  fact  (which  the  writer,  as  far  as  he 
can  discover,  was  the  first  to  notice)  that  there  happens  to  be 
the  same  numerical  relation  between  an  ounce  (av.)  and  a  kilo- 
gram, as  there  is  between  a  cubic  foot  and  a  cubic  meter;  in 
short,  there  are  35.26  ounces  (av.)  in  a  kilogram,  and  35.31 
cubic  feet  in  a  cubic  meter.  The  difference  is  only  one-seventh 
of  one  per  cent.,  which  can  be  ignored,  and  we  can  therefore  say 
that  if  the  relative  weights  in  an  equation  are  called  ounces 
(av.),  each  molecule  of  gas  in  the  equation  represents  22.22 
cubic  feet. 

Example. — The  production  of  acetylene  from  calcium  carbide : 

i 

CaC2  +  H2O  =  CaO  +  C2H2 
64   +  18  =  56  +   26 

Interpreting  by  weights,  and  calling  the  relative  weights 
ounces,  we  can  call  the  I  molecule  of  C2H2  gas  22.22  cubic  feet, 
so  that,  theoretically,  64  ounces  of  pure  carbide,  acting  on  18 
ounces  of  water,  produce  56  ounces  of  lime  and  26  ounces  of 
C2H2  gas,  the  volume  of  which  is  22.22  cubic  feet. 

WEIGHTS  AND  VOLUMES  OF  GASES. 

The  weight  of  one  cubic  meter  of  dry  air,  under  standard  con- 
ditions (at  0°  Centigrade  and  at  a  pressure  of  760  millimeters 
of  mercury),  is  1.293  kilograms.  The  composition  of  air  is: 

By  Weight.        By  Volume. 

Oxygen 3  21 

Nitrogen ,.        10  80 

or,  in  percentages, 

Oxygen 23.1  20.8 

Nitrogen 76.9  79.2 

While  these  may  not  represent  the  absolutely  accurate  aver- 
age composition  of  dry  air,  yet  the  variations  are  such  that  the 
above  simple  ratios,  3  to  10  and  21  to  80,  are  close  enough  for 
all  practical  purposes  in  metallurgy. 

The  weight  of  one  cubic  foot  of  dry  air  is  1.293  ounces  (av.). 

The  weight  of  one  cubic  meter  of  hydrogen  gas,  at  standard 
conditions,  is  0.09  kilogram  (1  cubic  foot,  0.09  ounces).  The 
formula  of  hydrogen  gas  is  H2,  is  molecular  weight  2;  and 
since  the  densities  of  all  gases  are  found  experimentally  to 


THE  CHEMICAL  EQUATION.  5 

be  proportional  to  their  molecular  weights,  it  follows  that  the 
density  of  any  gas  referred  to  hydrogen  is  expressed  numer- 
ically by  one-half  its  molecular  weight.  But  the  weight  of  a 
cubic  meter  of  gas  is  the  weight  of  a  cubic  meter  of  hydrogen 
multiplied  by  the  density  of  the  gas  referred  to  hydrogen;  thus, 
is  obtained  the  weight  of  a  cubic  meter  of  any  gas  whose  formula 
is  known.  Examples  follow: 

Molecular     Density  Referred     Weight  of  1 
Formula.  Weight.          to  Hydrogen.       Cubic  Meter. 

Hydrogen , H2  2  1  0.09  kilos. 

Water  vapor H2O  18  9  0.81  " 

Nitrogen N2  28  14  1.26  " 

Oxygen O2  32  16  1.44  " 

Carbon  monoxide.  .  .CO  28  14  1.26  " 

Carbon  dioxide CO2  44  22  1.98  " 

Marsh  gas CH4  16  8  0.72  " 

Etc.,  etc. 

In  the  case  of  water  vapor,  a  particular  explanation  is  neces- 
sary. It  cannot  exist  under  standard  conditions,  but  condenses 
to  liquid  at  100°  C.  if  under  760  millimeters  pressure.  It  does 
exist  at  lower  temperatures  than  100°,  but  only  under  partial 
pressures  of  fractions  of  an  atmosphere;  thus,  at  a  pressure  of 
1-50  atmosphere  (when  it  forms  1-50  of  a  mixture  of  gases) 
it  can  exist  un condensed  at  ordinary  temperatures  (15°  C.  or 
60°  P.).  The  above  weight  for  a  cubic  meter  of  water  vapor 
(0.81  kilos,  per  cubic  meter  at  standard  conditions)  is,  there- 
fore, only  a  hypothetical  value,  but  it  is  extremely  useful,  be- 
cause it  enables  us  to  calculate,  by  the  principles  to  be  ex- 
plained further  on,  the  weight  of  a  cubic  meter  of  water  vapor 
under  any  conditions  of  temperature  and  pressure  at  which  it 
is  possible  for  it  to  exist. 

CORRECTIONS  FOR  TEMPERATURE. 

The  volumes  of  all  permanent  gases  increase  uniformly  for 
uniform  increase  of  temperature,  so  that,  starting  with  a  given 
volume  at  0°  C.,  it  is  found  that  their  volume  increases  1/273 
for  every  degree  Centigrade  rise  in  temperature.  Thus,  at  273° 
C.,  the  volume  is  just  double  the  volume  at  0°.  Stating  this 
fact  in  another  way,  we  may  say  that  the  gas  acts  as  if  it  would 
have  no  volume  at  —  273°  C.,  and  would  increase  uniformly 


6  METALLURGICAL  CALCULATIONS. 

in  volume  from  this  point  up  to  all  measurable  temperatures, 
the  increment  being,  for  each  degree,  1-273  of  the  volume 
which  the  gas  has  at  0°  C.  A  still  briefer  statement  is  that  the 
volume  of  a  gas  is  proportional  to  its  temperature  above  —  273° 
C.,  or  to  its  absolute  temperature  —  the  latter  being  its  tem- 
perature in  C°  +  273. 

The  converse  of  these  principles  is,  that  the  density  of  a  gas; 
that  is,  the  weight  of  a  unit  volume,  varies  inversely  as  its 
absolute  temperature. 

In  Fahrenheit  degrees,  we  can  say  that  a  gas  expands  1-490 
(1-273x5-9)  for  every  degree  rise  above  32°  C.  ;  or  that  the 
volume  is  proportional  to  the  absolute  temperatures,  i.e.,  to 
the  F°.  -32  +  490  (  =  F.°  +  458). 

These  principles  are  in  constant  use  in  metallurgical  calcula- 
tions. Thus,  one  kilogram  of  coal  will  need  8  cubic  meters  of 
air  to  burn  it,  at  0°  and  760  millimeters  pressure.  What 
volume  will  that  be  at  30°  C.  and  the  same  pressure?  Since 
30°  C.  is  30  +  273  =  303°  absolute,  the  two  temperatures  will 
be  273  and  303,  and  the 

303 
Volume  at  30°  C.  =  volume  at  0°  C.X 


It  is  always  to  be  recommended  to  make  such  calculations  in 
the  above  form;  that  is,  to  first  put  down  the  known  volume, 
and  then  to  multiply  it  by  a  fraction,  the  numerator  and  de- 
nominator of  which  are  the  two  absolute  temperatures.  A 
moment's  reflection  will  show  which  way  the  fraction  must  be 
written;  if  the  new  volume  must  be  greater  than  the  old,  the 
value  of  the  fraction  must  be  greater  than  unity,  the  higher 
temperature  must  be  in  the  numerator;  if  the  fraction  were 
inverted,  we  know  that  the  result  would  be  less  than  the  start- 
ing volume  instead  of  greater,  which  would  be  wrong. 

Taking  an  example  in  Fahrenheit  degrees:  What  is  the  vol- 
ume under  standard  conditions  of  175  cubic  feet  of  gas  meas- 
ured at  90°  F.  and  standard  pressure  (29.93  inches  of  mer- 
cury)? Since  32°  F.  is  490°  absolute  and  90°  F.  is  548°  abso- 
lute, and  the  new  volume  must  be  less  than  the  starting  vol- 
ume, we  have 

Volume  at  32°  F,  =  175x         =  156.5  cubic  feet 


THE  CHEMICAL  EQUATION,  7 

CORRECTIONS  FOR  PRESSURE. 

The  principle  is  that  the  volumes  of  a  gas  are  inversely  as 
the  pressure  upon  it,  so  that  doubling  the  pressure  halves  the 
volume,  etc.  Since  the  practical  problems  almost  always  pre- 
sent the  pressure  as  two  numbers,  all  that  is  necessary  is  to 
multiply  the  original  volume  by  a  fraction  whose  numerator 
and  denominator  are  the  two  pressures  concerned,  and  ar- 
ranged with  the  numerator  the  larger  or  the  smaller  of  the  two 
numbers,  according  as  to  whether  the  final  volume  should  be 
greater  or  less  than  the  starting  one.  Putting  the  solution  in 
this  manner  avoids  the  primary  school  method  of  making  a 
proportion,  which  is  so  apt  to  be  expressed  upside  down,  and 
absolutely  avoids  error  with  the  minimum  exercise  of  brain 
power. 

Examples.  —  What  is  the  volume  of  100  cubic  meters  of  any 
gas,  if  the  pressure  is  changed  to  700  millimeters? 

760 
Answer:  100  X  z       =  108.6  cubic  meters. 


What  is  the  volume  at  standard  pressure  of  150  cubic  feet  of 
gas  measured  at  28.50  inches  of  mercury? 

oo    en 

Answer:  150  X^-^  =  142.8  cubic  feet. 

Z\y  .  y«5 

CORRECTIONS  FOR  TEMPERATURE  AND  PRESSURE. 

These  can  be  both  allowed  for,  by  simply  correcting  first  for 
one,  and  then  for  the  other.  Actually,  the  simplest  statement 
is  to  put  down  the  original  volume,  then  to  multiply  it  by  one 
fraction,  which  corrects  for  temperature,  and  again  by  another 
fraction  correcting  for  pressure,  thinking  out  carefully  for  each 
fraction  the  proper  way  of  expressing  it,  i.e.,  whether  it  should 
increase  or  decrease  the  volume. 

Examples.  —  What  does  100  cubic  meters  of  air  at  standard 
conditions  become  at  50°  C.  and  780  millimeters  pressure? 


en  i  070 

Solution:   100X     I*     X~  =  115.3  cubic  meters. 

to(J 


What  is  the  weight  of  one  cubic  meter  of  hydrogen  at  1000 


8  METALLURGICAL  CALCULATIONS. 

C.  and  250  millimeters  pressure,  its  weight  at  standard  condi- 
tions being  0.09  kilograms? 

273  250 

Solution:  0.09X  1000  +  273X76Q  =  °-00637  kil°Srams- 

What  weight  of  oxygen  is  in  1500  cubic  feet  of  dry  air  at 
100°  F.  and  at  28.50  inches  of  mercury?  (Refer  to  weight  of 
air  at  standard  conditions,  and  percentage  composition.) 


o 

Solution:  l^gSX  —  XXx1       X  1500  =  374  ounces. 

lo      OOo      Z\).\)o 

What  is  the  weight  of  50  cubic  meters  of  water  vapor  at  a 
temperature  of  30°  C.  and  a  pressure  of  31.6  millimeters? 

070          01      fi 

Solution:  0.81X^X^~X50  =  1.517  kilograms. 


070      qi    a 

or  50X^X^—^X0.81  =  1.517  kilograms. 
oOo       /OU 

The  first  expression  calculates  the  weight  of  a  cubic  meter 
of  water  vapor  at  the  assumed  conditions,  and  multiplies  by 
50;  the  second  calculates  the  hypothetical  volume  of  the  50 
cubic  meters  if  reduced  to  standard  conditions,  and  multiplies 
by  the  hypothetical  weight  of  a  cubic  meter  at  those  condi' 
tions. 

Problems  Illustrating  Preceding  Principles. 

Problem  1. 

A  bituminous  coal  contains  on  analysis: 
Carbon  ...............  73.60         Moisture  .............     0.60 

Hydrogen  ............     5.30        Ash  .................     8.05 

Nitrogen  .............      1.70 

Sulphur  ..............     0.75  100.00 

Oxygen  ..............   10.00 

It  is  powdered  and  blown  into  a  cement  kiln  by  a  blast  of  air. 

Required:  1.  The  volume  of  dry  air,  at  80°  F.  and  29  inches 
barometric  pressure  theoretically  required  for  the  perfect  com- 
bustion of  one  pound  of  the  coal. 

2.  The  volume  of  the  products  of  combustion,  using  no  excess 


THE  CHEMICAL  EQUATION.  9 

of  air,  at  550°  F.  and  29  inches  barometer,  and  their  percentage, 
composition. 

Solution:  The  reactions  of  the  combustion  are: 

C  +  O2  =  CO2  S  +  O2  =  SO2 

12     32       44  32     32       64 

2FP  +  O2  =  2H2O 
4     32       36 

Requirement  (1): 

The  oxygen  required  for  burning  one  pound  of  coal  is: 
Oxygen  for  carbon  ........  =  0.7360x32/12  ••=  1.9(53  pounds. 

Oxygen  for  hydrogen  ____      -=  0.0530x32/4     =  0.424 

Oxygen  for  sulphur  ......  =  0.0075X32/32  =  0.0075      " 

Total  required  .........................  ......  2.3945      " 

Oxygen  in  coal  ..............................  0.1000      " 

Oxygen  to  be  supplied  .......................  2.2945      " 

Nitrogen  accompanying  .......................  7.6483 

Air  necessary  .......  .  ........................  9.9428 

=  159.08  ounces  (av.). 
Volume  of  air  necessary  (standard  conditions) 

no 

=  123.03  cubic  feet. 


Volume  of  air  necessary  at  80°  F.  and  29  inches  barometer  = 
?^  =  139.4  cubic  feet.  (1) 


Requirement  (2):  Pounds. 

The  weight  of  CO2  formed  is  ...........  0.7360+1.963  =  2.699 

The  weight  of  H2O  formed  is  ...........  0.0530  +  0.424  =  0.477 

The  weight  of  moisture  is  ..........  ....  0.006 

The  weight  of  SO2  formed  is  ............  0.0075  +  0.0075  =  0.015 

The  weight  of  nitrogen  altogether  is  7.6483  +  0.0170  = 
7.6653  pounds.  .Converting  these  weights  into  ounces,  and 
dividing  each  by  the  weight  of  a  cubic  foot  of  each  gas  in  ounces, 
we  have  the  volume  of  these  theoretical  products  -at  standard 
conditions  : 


10  METALLURGICAL  CALCULATIONS. 

Volume  CO2  -  2.699x16+198  = 

43.184+  1.98  =    21.80  cubic  feet. 
Volume  H2O  -  0  483  X  16  +  0.81  = 

7728  +  0.81.  =  .     954      " 
Volume  SO2  =  0015x16  +  2.88  = 

02400  +  2.88  -      0.08      " 
Volume  N2  -  7665X16  +  1.26  = 

122.645  +  1.26=      9734      " 

Total  volume  at  standard  conditions.  ....  =    128.76 
Volume  at  550°  F.  and  29  inches  barometer  = 

-  2m  "        " 


. 

The    percentage    composition   by    volume   follows   from   the 
above  volumes  as: 

CO2  ...........  17.0  per  cent.         N2  ..........  75.5  per  cent. 

H20  .......  ...   7.4  —  rr 

SO2  ...........  0.1 

Problem  2. 

Natural  gas  in  the  Pittsburg  district  contains: 

Marsh  gas  .................  CH4  60.70  per  cent. 

Hydrogen,  ...............     H2  2903 

Ethane  ...................  C2H6       7.92 

Olefiant  gas  .............  .  .C2H<       0.98 

Oxygen  ..................  .O2       0.78 

Carbonous  oxide  ............  CO       0.58 

Required: 

(1)  The  volume  of  air  necessary  to  burn  it. 

(2)  The  volume  of  the  products  of  combustion. 

Reactions  : 

CH'  +  2O?  =    CO2  +  2H2O 

2H?  f   O2  =  2H2O 
2C2H6  +  7O-  =  4CO?  +  6H2O 

C2H4  +  3O2  =  2CO?  +  2H2O 

2CO  -f   O2  =  2CO2 
Solution  : 

Oxygen  required  for  CH<  .....   =  0.6070X   2    =  L2140  parts. 
Oxygen  required  for  H2  ......  =  0.2903  X   i     =--  0  1451      " 


THE  CHEMICAL  EQUATION.  11 

Oxygen  required  for  C2H6.  .  .  .  =  0.0792x7/2  =  0.2772  parts. 
Oxygen  required  for  C2H4  ____  =  0.0098  X   3     =  0.0294      " 
Oxygen  required  for  CO  ......  =  0.0058  X    i     =  0.0029      " 

1.6686      " 
Deduct  oxygen  already  present  ................  0.0078 

Leaves  oxygen  to  be  supplied  ................  1.6608 

Corresponding  to  air  ................  ,'    "I  -   =   7.985      "     (1) 

U  . 


Volumes  of  products  of  combustion: 

CO2  H2O  N2 

From  CH4  ...........  0.6070         1.2140 

From  H2  .........  ..  ..  0.2903 

From  C2H6  ...........  0.1584         0.2376 

From  C2H4  ...........  0.0196         0.0196 

From  CO  ............  0.0058 

From  air..  6.3242 


Total  products .  .  .  0 . 7908         1 . 761 5         6 . 3242    (2) 

The  above  solution  is  entirely  in  relative  volumes,  which 
may  be  all  considered  cubic  feet  or  cubic  meters,  and  are  true 
for  equal  conditions  of  temperature  and  pressure. 

Problem  3. 

A  Bessemer  converter  contains  10  metric  tons  of  pig  iron 
of  the  following  composition: 

Carbon 3 . 00  per  cent. 

Manganese 0.50 

Silicon 1.50 

Iron... 95.00 

On  being  blown,  one-third  the  carbon  burns  to  CO2,  the  rest 
to  CO;  5  per  cent,  of  iron  is  oxidized,  and  no  free  oxygen  escapes 
from  the  converter.  Blast  is  assumed  to  be  dry. 

Requirements : 

(1)  What  weight  of  oxygen  is  needed  during  the  blow. 

(2)  How  many  cubic  meters  of  air,  at  standard  conditions, 
will  be  needed. 

(3)  What  will  be  the  average  composition  of  the  gases. 


12  METALLURGICAL  CALCULATIONS 

Reactions  : 

C  +  O2  =  CO1  Si-fO'  -  SiO2 

12     32       44  28     32       60 

2C  +  O2  =  2CO  2Fe  +  O*  =  2FeO 

24     32       56  112     32       144 

2Mn  +  O2  =  2MnO 
110     32       142 

Oxygen  needed: 

C  to  CO2.  .    .  .  100  kilos.  X  32/12    =  266.7  kilos. 
CtoCO  .....  200       '     X32/24    =2667     " 
Mn  to  MnO.  .   50     "     X  32/1  10  =     14.5     " 
Si  to  SiO2    .  .150     "     X  32/28    =  171.4     " 
Fe  to  FeO.  .  .500     "     X32/112  --=  142.8     " 

Total  ...............................   862.1      "  (1) 

Nitrogen  accompanying  this  .........  =2873.7 

Air  needed  ...........................  3735  8     " 

3735  8 
Volume  of  air  ......  =  ~f~293'  =  2889'3  cublc  meters         (2) 

Volume  of  products  of  combustion: 


)       7 
CO2  =  1004266.7  =  366.7  kilos  =       -~   =  185.2  cu.  m. 


CO   »  200  +  266.7  --=  466.7  kilos  =  =  370.4  cu.  m. 

1  .zo 


N2  .  ,  2280.7      " 

Total  volume  .............................  2836.3      " 

Percentage  composition  by  volume: 

CO2  ..........................   6.5  per  cent. 

CO  ..........................  13.1 

N2  .  -.80.4 


CHAPTER  II. 
THE  APPLICATIONS  OF  THERMOCHEMISTRY. 

The  ordinary  interpretation  of  the  chemical  equation  by 
weight  gives  us  the  quantitative  relations  governing  the  re- 
actions of  substances  upon  each  other,  when  the  reaction  pro- 
ceeds to  a  finish.  Unfortunately,  much  chemical  instruction, 
as  given  in  our  elementary  schools,  and  even  in  some  of  the 
higher  ones,  stops  with  the  consideration  of  the  weight  re- 
lations and  does  not  proceed  to  those  equally  important  rela- 
tions, the  energetics  of  chemical  reactions.  In  most  of  the 
reactions  which  occur  in  practical  metallurgy,  the  quantity  of 
the  combustible  used,  or,  more  broadly,  the  amount  of  energy 
in  the  form  of  heat  or  electrical  energy,  necessary  for  pro- 
ducing the  reactions  desired,  is  the  controlling  factor  regulat- 
ing the  practicability  or  impracticability,  the  commercial  success 
or  failure,  of  the  process. 

The  relative  values  of  fuels,  the  manufacture  and  utilization 
of  gas,  the  principles  of  the  regenerative  furnace,  the  Bessemer 
process,  electric  reduction,  and  a  host  of  metallurgical  pro- 
cesses, depend  essentially  on  the  realization  and  utilization 
of  chemical  energy,  and  the  only  way  to  become  conversant 
with  the  amounts  of  energy  involved  or  evolved  in  these  opera- 
tions is  to  understand  thoroughly  the  thermochemistry  of  the 
reactions  concerned. 

THERMOCHEMICAL  NOMENCLATURE 

The  heat  evolved  when  compounds  are  formed  from  the 
elements  (in  a  few  cases  heat  is  absorbed)  is  determined  ex- 
perimentally by  the  use  of  the  calorimeter.  This  branch  is 
sometimes  called  "chemical  calorimetry ;  "  it  is  practically  a 
department  of  experimental  physics.  The  data  obtained  give 
the  heat  evolved  for  the  total  change  from  the  components 
at  room  temperature  to  the  resulting  products,  at  room  teir- 

13 


14  METALLURGICAL  CALCULATIONS. 

perature  (or  very  near  to  it),  and  may  be  expressed  per  unit 
of  weight  of  either  component  or  of  the  substance  formed. 
Thus,  if  carbon  is  burned  in  a  calorimeter  to  carbonic  acid 
gas,  CO2,  the  heat  evolved  may  be  reported  or  expressed  as 

8100  gram  calories  per  gram  of  carbon  burnt. 
Or,  3037  gram  calories  per  gram  of  oxygen  used. 
Or,  2209  gram  calories  per  gram  of  CO2  formed. 

Of  these  three  methods  of  expressing  the  results,  the  first 
is  the  more  often  used,  especially  by  the  physicist. 

The  chemist,  however,  finds  it  often  more  convenient  and 
logical  to  express  these  heats  of  combination  per  formula 
weight  of  the  substances  combining  and  of  product  formed. 
E.g.,  In  the  case  of  CO2,  which  contains  12  parts  of  car- 
bon and  32  of  oxygen  in  44  of  the  gas,  the  chemist  would  write, 

(C,  O2)  =  97,200, 

meaning  thereby  that  when  12  grams  of  carbon  is  burnt  by  32 
grams  of  oxygen,  forming  44  grams  of  carbon  dioxide,  there  is 
evolved  97,200  gram-calories.  These  are  the  laboratory  units; 
for  practical  purposes,  we  call  the  weights  kilograms  and  the 
heat  units  kilogram -calories  (gram-calories  are  abbreviated  to 
"  cal.  ";  kilogram -calories  to  "  Cal.").  Ostwald,  in  his  thermo- 
chemical  tables  writes  (C,  O2)  =  9721  K,  where  K  represents  a 
unit  100  times  as  large  as  a  gram-calorie,  if  weights  are  taken 
as  being  grams.  Berthelot  writes  (C,  O2)  =  97.2,  where  the 
heat  units  are  kilogram-calories,  if  the  weights  concerned  are 
taken  as  grams.  Both  these  methods  of  expression  are  liable 
to  cause  confusion ;  the  writer  prefers  to  follow  the  older  thermo- 
chemists  (Hess,  Naumann)  and  to  use  the  larger  number,  e.g., 
97,200,  which  then  means  gram-calories,  if  weights  are  taken 
in  grams  (laboratory  units),  and  kilogram-calories,  if  weights 
are  taken  in  kilograms  (practical  units). 

If  it  is  desired  to  work  in  "  British  Thermal  Units  "  (1  B.  T. 
U.  is  the  heat  needed  to  raise  one  pound  of  water  one  degree 
Fahrenheit),  the  weights  represented  by  the  formula  may  be 
called  pounds,  and  then  the  expression  is  as  follows: 

(C,  O2)  =  [97,200X9/5]  =  174,960  B.  T.  U. 
The  factor  9/5  is  simply  the  relation  of  1°  C.  to  1°  F. ;  the 


APPLICATIONS  OF  THERMOCHEMISTRY.  15 

reasoning  is,  of  course,  that  if  the  combustion  of  12  kilograms 
of  carbon  evolves  97,200  kilogram  calories,  or  would  heat 
97,200  kilograms  of  water  1°  C.,  that  the  combustion  of  12 
pounds  of  carbon  would  heat  97,200  pounds  of  water  1°  C.,  or 
174,960  pounds  1°  F.  From  the  equation  as  thus  expressed 
and  interpreted,  the  heat  of  combination  per  pound  of  carbon 
burnt,  or  of  oxygen  used,  or  of  product  formed,  may  be  found 
in  B.  T.  U.  by  dividing  174,960  by  12,  32  or  44,  respectively. 

Since  it  is  very  inconvenient,  as  well  as  unscientific,  to  have 
the  two  unrelated  heat  units,  with  their  resulting  double  sets 
of  experimental  data,  I  strongly  recommend  the  use  of  the 
metric  data  and  metric  Centigrade  heat  unit.  It  is,  however, 
sometimes  convenient,  when  all  the  data  of  a  problem  are 
given  in  English  weights,  to  use  as  the  unit  of  heat  the  "  pound 
— 1°  C.,"  or  the  heat  required  to  raise  the  temperature  of  one 
pound  of  water  1°  C.  This  may  be  called  the  "  pound  cal.," 
as  distinguished  from  the  B.  T.  U.  The  advantage  of  using  it 
is  that  all  the  experimental  data  of  the  metric  system  units 
are  at  once  transferable  to  the  English  weights.  E.g.,  (C,  O2) 
=  97,200  pound  cal.,  if  the  weights  concerned  in  the  formula 
(12,  32,  44)  are  called  pounds. 

The  thermochemist  gives  all  his  experimental  data  in  the 
form  above  explained,  and  we  will  now  give  all  the  important 
thermochemical  data  known  which  are  useful  in  metallurgical 
calculations,  the  data  being  for  the  reactions  beginning  and  ending 
at  15°  C.  (60°  F.): 

INTRODUCTION  TO  THERMOCHEMICAL  TABLES. 

In  the  thermochemical  tables  following,  the  heats  of  formation 
or  combination  are  given  first  for  the  molecular  weights  repre- 
sented by  the  formulas.  What  these  molecular  weights  are,  can 
be  found  by  adding  up  the  atomic  weights  of  the  atoms  repre- 
sented in  the  formula.  The  atomic  weight  table  on  pages  1  and 
2  supplies  the  necessary  data.  The  formula  given  represents, 
thermochemically,  the  union  of  the  constituent  parts  separated 
by  commas,  the  whole  reaction  being  enclosed  in  parentheses, 
to  distinguish  it  from  the  ordinary  chemical  formula.  Thus, 
while  (Ca,  0)  represents  the  union  of  40  parts  of  calcium  with 
16  parts  of  oxygen  to  form  56  parts  of  calcium  oxide,  and  (Ca, 
Si,  Oa)  represents  the  combination  of  40  parts  of  calcium  with 


16  METALLURGICAL  CALCULATIONS. 

28  parts  of  silicon  and  48  parts  of  oxygen  to  form  116  parts  of 
calcium  silicate,  (CaO,  SiO2)  represents  the  combination  of  56 
parts  of  lime  with  60  parts  of  silica  to  form  116  parts  of  calcium 
silicate. 

In  order  to  facilitate  the  use  of  these  data  in  metallurgical 
calculations,  the  succeeding  columns  give  the  heat  evolution 
per  unit  weight  of  the  metal  or  base  involved,  of  the  acid  element 
or  acid  radical  involved,  and  of  the  compound  formed.  This 
will  save  much  calculation,  for  while  the  heat  evolution  per 
formula  weight  is  of  prime  use  in  discussing  the  heat  energy  of 
chemical  reactions,  the  heats  of  combination  per  unit  weight  of 
constituents  or  of  the  product  are  most  convenient  in  ordinary 
calculations. 

The  column  headed  "To  Dilute  Solution"  gives  the  total  heat 
evolution  inclusive  of  the  heat  of  solution  in  a  large  excess  of 
water.  The  differences  between  corresponding  columns  under 
this  heading  and  the  heading  "Anhydrous,"  are  the  heat  of  solu- 
tion, per  formula  weight,  per  unit  of  base,  of  acid,  or  of  compound. 

The  order  in  which  the  elements  are  arranged  in  each  table  is 
the  order  of  their  heats  of  combination  with  unit  weight  of  the 
acid  element  or  radical,  which  expresses  the  real  order  of  affinity 
of  the  metals  in  each  case.  The  order  is  given  for  the  "An- 
hydrous" condition,  because  this  is  more  frequently  concerned 
in  metallurgy  than  "To  Dilute  Solution."  Theoretically,  the 
order  of  arrangement  "To  Dilute  Solution"  is  the  more  uniform 
and  logical,  besides  being  invariable,  but  it  is  less  useful  in  metal- 
lurgy, except  where  dilute  solutions  are  concerned. 

In  order  that  an  estimate  may  be  made  of  heats  of  formation 
so  far  undetermined,  there  is  given  at  the  end  of  the  tables  a 
list  of  the  " Thermochemical  Constants  of  the  Elements"  per 
chemical  equivalent  or  per  unit  of  valence  with  which  they  enter 
into  combination.  These  are  based  on  the  known  law  that  the 
heats  of  formation  of  salts  to  dilute  solution  are  additive  functions, 
being  simply  the  sum  of  the  thermochemical  combining  power 
of  the  basic  element  and  that  of  the  acid  element  or  radical. 
Assuming  arbitrarily  that  hydrogen  has  zero  for  its  thermo- 
chemical constant,  the  constants  for  those  elements  can  be  cal- 
culated for  which  any  thermochemical  data,  even  a  single  heat 
of  formation,  exist.  Thus,  referring  to  the  table,  on  page  38, 
57,200  for  Na  means  that  23  parts  of  sodium  going  into  combina- 


APPLICATIONS  OF  THERMOCHEMISTRY.  17 

tion  contributes  57,200  calories  towards  the  heat  of  formation 
of  the  compound  formed,  irrespective  of  what  it  is  combining 
with.  Similarly,  39,400  for  Cl  means  that  35.5  parts  of  chlorine 
going  into  combination  contributes  39,400  calories  towards  the 
heat  of  formation  of  any  chloride,  irrespective  of  the  element  it 
is  combining  with.  The  sum  of  these,  96,600,  is  the  heat  of 
formation  of  (Na,  Cl)  to  dilute  solution.  By  means  of  the  table 
of  constants  given,  simple  addition  gives  the  heat  of  formation 
of  a  large  number  of  salts,  from  their  elements  (taken  from  their 
usual  physical  condition  at  room  temperature)  per  chemical 
equivalent  of  base  and  acid,  or  of  compound,  involved.  This  is 
not  per  formula  weight,  except  for  monovalent  elements,  but  is 
per  chemical  equivalent  weight;  it  must  be  multiplied  by  the 
valency  of  the  base  in  the  formula,  that  is,  by  the  number  of 
chemical  equivalents  represented  by  the  formula,  to  get  the  heat 
of  formation  per  formula  weight.  Thus,  ;HjAl  40,100  plus 
Cl  39,400  gives  79,500  as  the  heat  of  formation  of  J^Al  C13; 
from  which  we  have  238,500  as  the  heat  of  formation  of  Al  CU, 
per  formula  weight. 

While  this  table  has  considerable  uses,  it  does  not  give  the  heats 
of  formation  of  anhydrous  compounds;  its  values  must  be  cor- 
rected by  the  heats  of  solution,  where  known,  to  give  the  values 
for  the  anhydrous  condition.  Yet,  these  are  frequently  correc- 
tions of  a  minor  order,  and  leave  the  table  highly  useful  for 
getting  first  approximations  to  undetermined  thermochemical 
data.  Names  of  some  elements  are  introduced  whose  thermo- 
chemical constants  are  unknown,  being  placed,  at  a  guess,  in 
their  most  probable  position  in  the  table.  All  quantities  are 
positive,  except  these  indicated  negative.  The  +  exponent 
after  a  symbol  represents  one  positive  valence,  the  —  exponent 
one  negative  valence.  A  more  detailed  explanation  of  the  matter 
can  be  found  in  an  article  by  the  writer  in  the  Trans.  American 
Electrochemical  Society,  1903,  Vol.  iv,  page  142. 


18 


METALLURGICAL  CALCULATIONS. 
HEAT  OF  FORMATION  OF  OXIDES 


Formula 

Anhydrous 

To  Dilute  Solution 

Molecu- 
lar 

Per  Unit  Weight  of 

Molecu- 
lar 

Per  Unit  Weight  of 

Metal 

Oxygen 

Oxide 

Metal 

Oxygen 

Oxide 

(Th,  02) 

326,000 

1,304 

10,188 

1,235 

(Mis,  OO1     . 

472,000 

1,655 

9,850 

1,417 

(La2,  Os) 

444,700 

1,602 

9,265 

1,364 

(Nd!f  03)      • 

435,100 

1,506 

9,006 

1,291 

(Mg,  0) 

143,400 

5,975 

8,963 

3,585 

148,800 

6,200 

9,300 

3,720 

(Pr«,  00 

412,400 

1,467 

8,592 

1,297 

(Li2,  0 

135,800 

9,700 

8,488 

4,527 

167,000 

11,929 

10,437 

5,567 

(Ba.O) 

133,400 

974 

8,338 

872 

161,500 

1,179 

10,938 

1,051 

(Ca.  0) 

131,500 

3,288 

8,219 

2,348 

149,600 

3,740 

9,350 

2,671 

(Sr,  0) 

131,200 

1,508 

8,200 

1,274 

158,400 

1,821 

9,900 

1,538 

(Ah.  OO 

392,600 

7,270 

8,179 

3,849 

(V.,  Oi) 

353,200 

3,463 

7,358 

2,355 

(Ce,  00 

'  224,600 

1,603 

7,019 

1,306 

(Ti,  00    , 

218,400 

4,550 

6.825 

2,730 

(U3,  00 

845,200 

1,184 

6,603 

1,004 

(V,  0) 

104,300 

2,045 

6,519 

1,557 

(U,  00 

303,900 

1,277 

6,341 

1,063 

(Na2,  0) 

100,400 

2,183 

6,275 

1,620 

155,900 

3,389 

9,744 

2,515 

(Kt.  0) 

98,200 

1,259 

6,138 

1,045 

165,200 

2,118 

10,325 

1,758 

(Si,  OO 

196,000 

7,000 

6,125 

3,267 

(Mn,  O) 

90,900 

1,653 

5,681 

1,280 

(B»,  OO 

272,600 

12,391 

5,679 

3,894 

279,900 

12,723 

5,831 

3,999 

(Vt,  Os) 

447,000 

4,324 

5,588 

2,423 

(Zr,  0.) 

177,500 

1,972 

5,547 

1,455 

(Zn,  0) 

84,800 

1,305 

5,300 

1,058 

(Mm,  OO 

328,000 

1.98S 

5,125 

1,434 

(Cra,  0.) 

243,900 

2,345 

5,081 

1,605 

(Li.,  0*) 

152,650 

10,904 

4,833 

3,318 

159,840 

11,417 

4,995 

3,474 

(P».  06) 

365,300 

5,895 

4,566 

2,572 

400,900 

6,466 

5,013 

2,823 

(Ba,  0.) 

145,500 

1,062 

4,547 

861 

(Mo,  O«) 

142,800 

1,488 

4,463 

1,116 

(Sn,  0). 

70,700 

599 

4,419 

527 

(Sn,  00 

141,300 

1,197 

4,416 

942 

(CO,  0)  (gas) 

68,040 

2,430 

4,253 

1,546 

73,940 

2,641 

4,621 

1,680 

[  solid 

70,400 

35,200 

4,400 

3,911 

(Hi,  0)  1  liquid 

69,000 

34,500 

4,313 

3,833 

I  gas 

58,060 

29,030 

3,629 

3,226 

(Pe,,  04) 

270,800 

1,612 

4,231 

1,167 

(Cd,  0) 

66,300 

592 

4,144 

518 

(Pe,  0) 

65,700 

1,173 

4,106 

913 

(W.  00 

131,400 

714 

4,106 

620 

(W,  00 

196,300 

1,067 

4,089 

846 

(Fes,  00 

195,600 

1,746 

4,075 

1,223 

(Co   0) 

64,100 

1,086 

4,006 

855 

(Mn,  OO 

125,300 

2,278 

3,916 

1,440 

Mi  =  Mischmetal — a    mixture    of   rare   earth    metals    sold   commercially,   containing 
cerium,  thorium,  yttrium,  etc. 


APPLICATIONS  OF  THERMOCHEMISTRY. 


19 


HEAT  OF  FORMATION  OF  OXIDES.     (Continued] 


Formula 

Anhydrous 

To  Dilute  Solution 

ifdte*- 

lor 

Per  Unit  Weight  of 

Molecu- 
lar 

Per  Unit  Weight  of 

Meal 

Oxygen 

Oxide 

Metal 

Oxygen 

Oxide 

(Ni,  0) 

61,500 

1.051 

3,844 

826 

(Naz,  02) 

119,800 

2,604 

3,744 

1,536 

(Mo,  Oa) 

175,000 

1,823 

3,646 

1,215 

(Sb«,  03) 

166,900 

695 

3,479 

580 

(Sbi,  O«) 

209,800 

874 

3,278 

690 

(Ast,  O3) 

156,400 

1,043 

3.258 

790 

148,900 

993 

3,102 

752 

(Pb,  0) 

50,800 

245 

3,175 

228 

(C,  Oi)  (gas) 

97,200 

8,100 

3.038 

2,209 

103,100 

8,592 

3,222 

2,343 

(Cr,  Oi) 

140,000 

2,692 

2,917 

1,400 

(Bit.  O3) 

139,200 

335 

2,900 

300 

(Sb»,  Os) 

231,200 

963 

2,890 

723 

(As«.  Oa) 

219,400 

1,463 

2,743 

954 

225,400 

1,503 

2,818 

980 

(CU2,   0) 

43,800 

344 

2,738 

306 

(Tl»,  0) 

42,800 

105 

2,675 

101 

39,700 

97 

.2,481 

—94 

(Te,  Oj) 

78,300 

624 

2,447 

497 

(Cu,  0) 

37,700 

593 

2,356 

474 

(S,  Oi) 

69,260 

2,196 

2,196 

1,098 

77,600 

2;425 

2,425 

1,213 

(Pb,  Oz) 

63,400 

306 

1,981 

265 

(S,  03)  (gas) 

91,900 

2,872 

1,915 

1,149 

141,000 

4,406 

2,938 

1.763 

(Tit,  Oi) 

87,600 

215 

1,825 

192 

(C,  0)  (gas) 

29,160 

2,430 

1,823 

1,041 

(H«,  Oi) 

47,300 

23,650 

1,478 

1.391 

(Hgi,  O) 

22,200 

56 

1,388 

53 

(Hg,  0) 

21.500 

108 

1,344 

100 

(Pd,  0) 

21,000 

198 

1,313 

172 

(Pt,  0) 

17,000 

87 

1,063 

81 

(Ag«.  0) 

7,000 

32 

438 

30 

(Am.  Oi) 

-11,500 

-29 

-240 

-26 

HEAT  OF  FORMATION  OF  HYDRATES 
A.  From  the  Elements 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(Li,  0,  H) 

112,200 

16,030 

118,000 

16,860 

(Mg,  02,  H2) 

217,800 

9,075 

(Sr,  02,  H2 

) 

217,300 

2,498 

227,400 

2,614 

(Ca,  0,,  H,) 

215,600 

5,390 

219,500 

5,488 

(K,  0,  H) 

104,600 

2,682 

117,100 

3,003 

(Na,  0,  H) 

102,700 

4,465 

112,450 

4,889 

(Al,  0,,  H3 

) 

301,300 

11,160 

(N,  H«,  0, 

H) 

88,800 

90,000 

5,294 

(Zn,  Oi,  HO 

151,100 

2,325 

solid 

70,400 

(H,  0,  H) 

liquid 

69,000 

gas 

58,060 

(Tl,  O,  H) 

57,400 

260 

54,300 

246 

(Bi,  0,,  H,) 

171,700 

825 

20 


METALLURGICAL  CALCULATIONS. 


HEAT  OP  FORMATION  OF  HYDRATES — Continued 
A.    From  the  Elements 


Formula 

Anhydrous 

To  Dilute  Solution 

and 
Reaction 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(Te,  04f  H2) 

166,740 

1,323 

(Te,  Olf  H2) 

145,600 

1,156 

(Se,03,  HO 

124,500 

1,576 

(Se,  04,  HO 

128,220 

1,623 

145,020 

1,836 

(Tl,  03,  HO 

78,700 

386 

B.  From  Metal,  Oxygen  and  Water 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 

Formula 
Weight 

Per  Unit 
Weight  of 
Metal 

Per  Unit 
Weight  of 
Oxygen 

Per 

Formula. 
Weight 

Per  Unit 
Weight  of 
Metal 

Per  Unit 
Weight  of 
Oxygen 

(Li2,  0,  H20) 

155,400 

11,100 

9,710 

167,000 

11,930 

10,440 

(Mg,  0,  H20) 

148,800 

6,200 

9,300 

(Sr,  O,  H2O 

148,300 

1,705 

9,280 

158,400 

1,822 

9,900 

(Ca,  0,  H20) 

146,600 

3,665 

9,162 

150,500 

3,763 

9,406 

(K2,  0,  H20) 

140,200 

1,797 

8,762 

165,200 

2,118 

10,325 

(Na2,  0,  H20) 

136,400 

2,965 

8,525 

155,900 

3,389 

9,744 

(A12,  03,  3H20) 

395,600 

7,326 

8,242 

(Zn,  0,  H20) 

82,100 

1,263 

5,131 

(T12,  0,  H20) 

45,800 

112 

2,863 

39,600 

97 

2,475 

(Bi2,  03,  3H20) 

136,400 

328 

2,842 

(Te,  02,  H20) 

76,600 

6,079 

2,394 

(Te,  03,  H20) 

97,740 

776 

2,036 

(Se,  02,  H20) 

55,500 

703 

1,735 

(Se,  O3,  H2O 

59,220 

750 

1,234 

76,020 

962 

1,583 

(T12,  O3,  3H2O) 

86,400 

212 

1,800 

C.  From  Metallic  Oxide  and  HZ0 


Anhydrous 

To  Dilute  Solution 

Formula 
and 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Reaction 

Formula 
Weight 

Metallic 
Oxide 

H20 

Hydrate 

Formula 
Weight 

Metallic 
Oxide 

Com- 
bined 

mo 

Dis- 
solved 
Hydrate 

(Li,  O,  H2O) 

19,600 

653 

1,089 

408 

31,200 

1,040 

1,733 

650 

(MgO,  H20) 

5,400 

135 

300 

93 

(SrO,  H2O) 

17,100 

166 

950 

141 

27,200 

264 

1,511 

225 

(CaO,  H»O) 

15,100 

270 

839 

204 

19,000 

339 

1,056 

257 

(K20,  H20) 

42,000 

447 

2,333 

375 

67,000 

713 

3,722 

598 

(Na20,  H20) 

36,000 

581 

2,000 

450 

55,500 

895 

3,083 

694 

(N  H4)20,  H'O 

(A1Z03,  3H20) 

3,000 

30 

56 

19 

(ZnO,  H2O) 

-2,700 

-33 

-150 

-27 

(ThO,  HzO) 

3,000 

7 

167 

7 

-3,200 

-7 

-178 

-7 

(BizOa,  3H20) 

-2,800 

*T 

-52 

-6 

(TltOi,  3H'0) 

-1,200 

-3 

-22 

-2 

APPLICATIONS  OF  THERMOCHEMISTRY. 


21 


HEAT  OF  FORMATION  OF  SULFIDES 


Formula 

Anhydrous 

To  Dilute  Solution  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Sulfur 

Sulfide 

Metal 

Sulfur 

Sulfide 

(Li2    S) 

115,400 

8,243 

3,606 

2,509 

\-L'12j    *-V 

(K,,  S) 

103,500 

1,327 

3,234 

941 

113,500 

1,455 

3,547 

10,319 

(Ba,  S) 

102,900 

751 

3,216 

609 

109,800 

802 

3,431 

474 

(Sr,  S) 

99,300 

1,141 

3,103 

835 

106,700 

1,225 

3,334 

897 

(Na,f  S) 

89,300 

1,941 

2,791 

1,145 

104,300 

2,267 

3,259 

1,337 

(Nd2,  S,) 

285,900 

993 

2,977 

744 

(Cat  S) 

94,300 

2,358 

2,947 

1,310 

100,600 

2,515 

3,144 

1,397 

(Mg,  S) 

79,400 

3,308 

2,481 

1,418 

(Mn,  S) 

45,600 

829 

1,425 

524 

(Zn,  S) 

43,000 

662 

1,344 

443 

(A12,  SO 

126,400 

2,341 

1,316 

843 

(N,  H5,  S) 

40,000 

2,105 

1,250 

776 

36,700 

1,931 

1,147 

720 

(Cd,  S) 

34,400 

307 

1,075 

239 

(K,  SO 

59,300 

1,521 

927 

576 

59,700 

1,531 

933 

579 

(Na,  S,) 

49,500 

2,152 

773 

556 

54,400 

2,365 

850 

611 

(B2,  S3) 

75,800 

3,445 

790 

642 

(Fe,  S) 

24,000 

428 

750 

273 

(Co,  S) 

21,900 

371 

685 

241 

(T12,  S) 

21,600 

106 

675 

92 

(Cu,f  S) 

20,300 

160 

634 

127 

(Pb,  S) 

20,200 

98 

631 

85 

(Si,  SO 

40,000 

1,429 

625 

435 

(Ni,  S) 

19,500 

333 

609 

215 

(Sb,,  SO 

34,400 

143 

358 

102 

(Hgf  S) 

10,600 

53 

331 

46 

(Cu,  S) 

10,100 

159 

316 

106 

(H2,  S) 

(gas) 

4,800 

2,400 

150 

141 

9,500 

4,750 

297 

2Y9 

(Ag2,  S) 

3,000 

14 

94 

12 

(C,  SO 

(gas) 

-25,400 

-2,117 

-397 

-334 

(liquid) 

-19,000 

-1,583 

-297 

-250 

(1,3) 

0 

0 

0 

0 

22 


METALLURGICAL  CALCULATIONS 


HEAT  OF  FORMATION  OF  SELENIDES    • 


Anhydrous 

To  Dilute  Solution 

Formula 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Weight 

Metal 

Selen- 
ium 

Selen- 
ide 

Wright 

Metal 

Selen- 
ium 

Selen- 
ide 

(Liz,  Se) 

83,000 

5,929 

1,051 

892 

93,700 

6,693 

1,186 

1,008 

(K2,  Se) 

79,600 

1,021 

1,008 

507 

87,900 

1,127 

1,113 

560 

(Ba,  Se) 

69.900 

510 

885 

324 

(Sr,  Se) 

67,600 

777 

856 

408 

(Na2.  Se) 

60,900 

1,324 

720 

487 

78,600 

1,709 

995 

629 

(Ca>  Se) 

58,000 

1,450 

727 

487 

(Zn,  Se) 

30,300 

466 

384 

210 

(Cd,  Se) 

23,700 

212 

300 

124 

(Mn,  Se) 

22,400 

407 

284 

167 

(N,  Hs,  Se) 

17,800 

937 

225 

184 

12,800 

674 

162 

136 

(Cu,  Se) 

17,300 

272 

219 

114 

(Pb,  Se) 

17,000 

82 

215 

59 

(Pe,  Se) 

15,200 

262 

,192 

111 

(Ni,  Se) 

14,700 

251 

186 

107 

(Cd,  Se) 

13,900 

236 

176 

101 

(Th,  Se) 

13,400 

33 

170 

21 

(Gu«,  Se) 

8,000 

63 

101 

39 

(Hg,  Se) 

6,300 

32 

80 

23 

(Ag2,  Se) 

2,000 

93 

25 

7 

(H2,  Se)  (gas) 

-25,100 

-12,550 

^-318 

-310 

-  15,800 

-7,900 

-200 

-195 

(N,  Se) 

-42,300 

-3,021 

-534 

,'-455 

HEAT  OF  FORMATION  OF  TELLURIDES 


Anhydrous 


Formula 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Tellurium 

Telluride 

(Zn,  Te) 

31,000 

477 

246 

162 

(Cd,  Te) 

16,600 

148 

132 

70 

(Co,  Te) 

13,000 

220 

103 

70 

(Fe,  Te) 

12,000 

214 

95 

66 

(Ni,  Te) 

11,600 

198 

92 

63 

(T12,  Te) 

10,600 

26 

84 

20 

(Cu,,  Te) 

8,200 

64 

65 

32 

(Pb,  Te) 

6,200 

30 

49 

19 

(H2,  Te)  (gas) 

-34,900 

-17,450 

-277 

-272 

APPLICATIONS  OF  THERMOCHEMISTRY. 


23 


HEAT  OF  FORMATION  OP  ARSENIDES 


Formula 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Arsenic 

Arsenide 

(H3, 

As)  (gas) 

-44,200 

-14,733 

-589 

-567 

HEAT  OF  FORMATION  OF  ANTIMONIDES 


Formula 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Antimony 

Antimonide 

(H,f 

Sb)  (gas) 

-86,800 

-28,933 

-723 

-706 

HEAT  OF  FORMATION  OF  PHOSPHIDES 


Per 

Per  Unit  Weight  of 

Formula 

Formula 

Weight 

Metal 

Phosphorus 

Phosphide 

(Mn3,  P,) 

70,900 

430 

1,144 

312 

(H3,  P)  (gas) 

4,900 

1,633 

158 

144 

(Fe,  P) 

±0 

±0 

±0 

±0 

HEAT  OF  FORMATION  OF  NITRIDES 


Anhydrous 

To  Dilute  Solution 

Formula 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Weight 

Metal 

Nitrogen 

Nitride 

Weight 

Base 

Acid 

Com- 
pound 

(Bus.  N2) 

149,400 

363 

5,336 

340 

(Mga,  N2) 

119,700 

1,662 

4,275 

1,197 

(Cas,  N2) 

112,200 

935 

4,007 

758 

(Li,,  N) 

49,500 

2,357 

3,536 

1,414 

(Al,  N) 

45,450 

1,683 

3,246 

1,109 

(Li,  N) 

18',750 

2,679 

1,339 

•893 

(Hs,  N)  (gas) 

12,200 

4.067 

871 

718 

21,000 

7,000 

1,500 

1,235 

(liquid) 

16,600 

5,533 

1,186 

976 

(P»,  N,) 

81,500 

876 

1,164 

500 

24 


METALLURGICAL  CALCULATIONS. 


HEAT  OF  FORMATION  OF  HYDRIDES 


Formula 

Per 

Formula 

Per  Unit  Weight  of 

Per 

rTOtflttlCL 

Per  Unit  Weight  of 

Weight 

Metal 

Hydro- 
gen 

Hy- 
dride 

Weight 

Base 

Acid 

Com- 
pound 

(Ca,  HO* 

46,200 

1,155 

23,100 

1,100 

(Li,  H) 

21,600 

3,086 

21,600 

2,700 

(Sr,  HO 

38.400 

441 

19,200 

431 

(Ba,  H») 

37,500 

274 

18,750 

271 

(Ptis,  H) 

16,950 

6 

16,950 

6 

(Ptio,  H) 

14,200 

7 

14,200 

7 

(Pu,  He) 

53,400 

144 

8,900 

.141 

(Pdu,  H) 

4,600 

3 

4,600 

3 

(P,  Hi)  (gas) 

+4,900 

158 

+  1,633 

144 

(Si,  H4)  (gas) 

-6,700 

-239 

-1,675 

-209 

(N4,   HO 

-19,000 

-339 

-4,750 

-317 

-26,100 

-466 

-6,525 

-435 

(As,  H3) 

-44,200 

-589 

-14,730 

-567 

(Sb,  HJ) 

-86,800 

-723 

-28,930 

-706 

HEAT  OF  FORMATION  OF  HYDROCARBONS 
(From  Amorphous  Carbon,  to  Gas,  Unless  Otherwise  Specified} 


Formula 
and 
Reaction 

Name 

State 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Carbon 

Hydro- 
gen 

Com- 
pound 

(C,  HO 

Methane  (Marsh  gas) 

Gas 

+21,700 

+  1,808 

+5,425 

+  1,356 

(C2,  HJ) 

Acetylene 

Gas 

-52,300 

-2,179 

-26,150 

-2,016 

(C2,  HO 

Ethylene  (olefiant  gas) 

Gas 

-8,700 

-363 

-2,175 

-311 

(C,,  He) 

Ethane  (ethylene  hydride) 

Gas 

+29,100 

+  1,213 

+4,850 

+970 

(Ca,  HO 

Allylene 

Gas 

-44,000 

-1,222 

-11,000 

-1,100 

(C3,  HO 

Propylene 

Gas 

-700 

-22 

-133 

-19 

(Cj.  HO 

Propane  (propylene  hydride) 

Gas 

+39,200 

+  1,089 

+4.900 

+891 

(Ca,  HO 

Benzene 

Gas 

+7,200 

+  100 

+  1,200 

+92 

Liquid 

+  14,400 

+200 

+  2,400 

+  185 

(Cr,  HO 

Toluol 

Liquid 

+  21,900 

+261 

+  2,738 

+238 

(Cio,  HO 

Naphthalene 

Liquid 

+  9,100 

+76 

+  1,138 

+  71 

Solid 

+  13,700 

+  114 

+  1,713 

+  107 

(Cio,  Hie) 

Turpentine 

Gas 

+23,800 

+  197 

+  1,488 

+  175 

Liquid 

+  33,200 

+277 

+2,075 

+244 

(Ci4.  Hio) 

Anthracene 

Solid 

-2,600 

-16 

-260 

-15 

(C,  Hi,'0) 

Methyl  alcohol  (wood  alcohol) 

Gas 

+56,000 

4,667 

14,000 

1,750 

Liquid 

+  65,050 

5,420 

16,260 

2.033 

(Ct.  He,  0) 

Ethyl  alcohol  (grain  alcohol) 

Gas 

+  64,200 

2,675 

10,700 

1.396 

Liquid 

+74,900 

3,120 

12,480 

1,628 

(C.,  He,  0) 

Acetone 

Gas 

+  63,150 

1,754 

10,525 

1,089 

Liquid 

+  71,300 

1,980 

11,880 

1,229 

APPLICATIONS  OF  THERMOCHEMISTRY. 


25 


HEAT  OF  COMBUSTION  OF  HYDROCARBONS 


Formula 

Name 

State 

Per  Formula  Weight 

Calor- 
ies per 

M3 

Calor- 
ies per 
Kg 

B.T.U. 

B.T.U. 

per 
/M 

per 
Ib. 

Calorime- 
ter Value 

Practical 
Value 

Water  formed  not  condensed 

CH4 
CiHj 
C,H4 
CiHe 
C3H4 
CaHe 
C3Hs 
CeH6 

CyHs 
CioHs 

CioH16 

CuHio 
CHiO 

C2H«O 
CaH.O 

Methane 
Acetylene 
Ethylene 
Ethane 
Allylene 
Propylene 
Propane 
Benzene 

Toluol 
Naphthalene 

Turpentine 

Anthracene 
Methyl  alcohol 

Ethyl  alcohol 
Acetone 

Gas 
Gas 
Gas 
Gas 
Gas 
Gas 
Gas 
Gas 
Liquid 
Liquid 
Liquid 
Solid 
Gas 
Liquid 
Solid 
Gas 
Liquid 
Gas 
Liquid 
Gas 
Liquid 

213,500 
315,700 
341,100 
372,300 
473.600 
499.300 
528,400 
783,000 
775,800 
934,500 
1,238,900 
1,234,300 
1,500,200 
1,490,800 
1,708,400 
179,200 
170,150 
337,200 
326,500 
435,450 
427,300 

191,620 
304,760 
319,220 
339,480 
451,720 
466,480 
484,640 
750,180 
742,980 
890,740 
1,195,150 
1,190,550 
1,412,680 
1,403,280 
1,653,700 
157,320 
148,270 
304,380 
293,680 
402,630 
394,480 

8,623 
13,714 
14,365 
15,277 
20,327 
20,992 
21,812 
33,758 

63,570 

11,976 
11,722 
11,400 
11,315 
11,295 
11,105 
11,015 
9,618 
9,525 
9,682 
9,337 
9,300 
10,385 
10,318 
9,290 

970 
1,543 
1,616 
1,719 
2,287 
2,362 
2,464 
3,798 

21,555 
21,000 
20,520 
20,365 
20,150 
19,995 
19,825 
17,312 
17,145 
17,430 
16,805 
16,740 
18,695 
18,570 
16,545 
8,854 
8,340 
11,910 
11,490 
12,495 
12,240 

7,152 

7,079 

4,919 
4,633 

786 

13,700 

6,617 
6,384 

1,541 

18,120 

6,942 
6,801 

2,038 

HEAT  OF  FORMATION  OF  CARBIDES 


Formula 

Per  Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Carbon 

Carbide 

(A14,  C.) 

232,000 

2,148 

6,445 

1,815 

(Si,  C) 

26,520 

947 

2,210 

663 

(Mn,,  C) 

9,900 

60 

825 

57 

(Fe3,  C) 

8,460 

50 

705 

47 

(Ca,  CO 

-7,250 

181 

302 

113 

(Na,  C) 

-4,400 

191 

367 

126 

(Li,  C) 

-5,750 

821 

479 

303 

(Ag,  C) 

-43,575 

403 

3,631 

363 

(N2,  CO 

-73,000  (gas) 

-2,607 

-3,042 

-1,404 

28 


METALLURGICAL  CALCULATIONS. 


HEAT  OF  FORMATION  OF  CARBONATES 
A.  From  the  Elements 


Formula 

Anhydrous 

To  Dilute  Solution 

Per  Formula    ' 
Weight 

Per  Unit 
Weight  of  Metal 

Per  Formula 
Weight 

Per  Unit 
Weight  of  Metal 

(Ba,  C,  0.) 

286,300 

2,090 

(K2,  C,  03) 

282,100 

3,617 

288,600 

3,700 

(Sr,  C,  03) 

281,400 

3,234 

(Ca,  C,  03) 

273,850 

6,846 

(Na2,  C,  03) 

273,700 

5,950 

279,300 

6,072 

(Mg,  C,  03) 

269,900 

11,245 

(Mn,  C,  03) 

210,300 

3,824 

(N,  H6,  C,  03) 

208,600 

10,980 

202,300 

10,640 

(Zn,  C,  03) 

197,500 

3,038 

(Pe,  C,  03) 

187,800 

3,354 

(Cd,  C,  03) 

183,200 

1,636 

(Pb,  C,  03) 

170,000 

821 

(Cu,  C,  03) 

146,100 

2,301 

(Ag2,  C,  03) 

123,800 

593 

B.  From  Metal,  Oxygen  and  C02 


Anhydrous 

To  Dilute  Solution 

Formula 
and 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Weight 

Metal 

Oxygen 

Weight 

Metal 

Oxygen 

(Ba,  0,  C02) 

189,100 

1,380 

(K2,  O,  CO2) 

184,900 

2,370 

191,400 

2,454 

(Sr,  0,  C02) 

184,200 

2,117 

(Ca,  0,  C02) 

176,650 

4,416 

(Na2,  O,  CO2) 

176,500 

3,837 

182,100 

3,959 

(Mg,  0,  C02) 

172,700 

7,195 

(Mn,  O,  CO2) 

113,100 

2,056 

(N,  H6,  O,  CO2) 

111,400 

5,863 

105,100 

5,532 

(Zn,  O,  CO2) 

100,300 

1,543 

(Fe,  0,  C02) 

90,600 

1,618 

(Cd,  0,  C02) 

86,000 

768 

(Pb,  O,  CO2) 

72,800 

352 

(Cu,  0,  C02) 

48,900 

770 

(Ag2,  0,  C02) 

26,600 

123 

APPLICATIONS  OF  THERMOCHEMISTRY. 

C.  From  Metallic  Oxide  and  COZ 


29 


Anhydrous 

To  Dilute  Solution 

Formula 

Per  Unit  Weight  of 

Per  Unit  Weight  of 

Reaction 

Per 

Formula 

Per 

Formula 

Weight 

Metallic 
Oxide 

CO2 

Car- 
bonate 

Weight 

Metallic 
Oxide 

COt 

Car- 
bonate 

(BaO,  CO2) 

55,700 

364 

1,266 

282 

(K2O,  CO2) 

86,700 

922 

1,970 

628 

93,200 

992 

2,118 

675 

(SrO,  CO*) 

53,000 

515 

1,205 

361 

(CaO,  CO2) 

45,150 

806 

1,026 

451 

(Na2O,  CO2) 

76,100 

1,219 

1,718 

713 

81,700 

1,318 

1,857 

771 

(MgO,  CO2) 

29.300 

733 

666 

349 

(MnO,  CO2) 

22,200 

313 

505 

193 

(NHs,  O,  CO2) 

22,600 

646 

514 

286 

16,300 

466 

370 

206 

(ZnO,  CO2) 

15,500 

191 

352 

124 

(FeO,  CO2) 

24,900 

346 

566 

215 

(CdO,  CO2) 

19,700 

154 

448 

115 

(PbO,  CO2) 

22,000 

99 

500 

83 

(CuO,  CO2) 

11,200 

140 

255 

91 

(Ag2O,  CO2) 

19,600 

82 

445 

71 

HEAT  OF  FORMATION  OF  BICARBONATES 
A.  From  the  Elements 


Anhydrous 

To  Dilute  Solution 

Formula 

and 
Reaction 

Per  Formula 
Weight 

Per  Unit 
Weight  of 
Metal 

Per  Formula 
Weight 

Per  Unit 
Weight  of 
Metal 

(K,  H, 

C,  Os) 

233,300 

5,987 

228,000 

5,846 

(Na,  H 

,  C,  Os) 

227,000 

9,870 

222,700 

9,683 

(N,  H« 

H,  C,  Os) 

205,300 



199,000 

B.  From  Metal,  Oxygen,  HZ0  and  COZ 


Formula  and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 

For- 
mula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metal 

CO2 

Metal 

C02 

Com' 
Pound 

(K2,  O,  H20,  2C02) 

203,200 

2,605 

2,309 

1,016 

192,600 

2,469 

2,189 

963 

(Na2,  O,  H2O,  2C02) 

190,600 

4,143 

2,166 

1,135 

182,000 

3,957 

2,036 

1.083 

C.  From  Metallic  Oxide,  HZ0  and  COZ 


Formula 

Anhydrous 

T<?  Dilute  Solution 

Per 
For- 
mula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weipht  of 

Metallic 
Oxide 

COt 

Com- 
pound 

Metallic 
Oxide 

CO,  ' 

Com- 
pound 

(K2O,  H2O,  2COt) 
(Na2O,  H2O,  2CO2) 

105,000 
90,200 

1,117 
1,961 

1,193 
1,025 

525 
537 

94,400 
81,600 

1,004 
1,774 

1,073 
927 

472 

486 

30 


METALLURGICAL  CALCULATIONS. 


HEAT  OF  FORMATION  OF  NITRATES 
A.  From  the  Elements 


Anhydrous 

To  Dilute  Solution 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(K,  N,  Os) 

119,000 

3,051 

110,700 

2,582 

(Ba,  N«,  Oe) 

227,200 

1,658 

217,800 

1,590 

(Li,  N,  O3) 

111,610 

15,945 

111,910 

15,985 

(Na,  N,  O3) 

110,700 

4,813 

106,000 

4,609 

(Sr,  N2,  Oe) 

219,800 

2,726 

214,700 

2,468 

(Ca,  N«,  Oe) 

202,000 

5,050. 

206,000 

5,150 

(Mg,  N»,  Oe) 





206,300 

8,596 

(Mn,  N«,  Oe) 





146,900 

2,671 

(Zn,  N«,  Oe) 





131,700 

2,022 

(Tl,  N,  03) 

58,150 

285 

48,180 

236 

(Pe,  N»,  Oe) 





119,000 

2,125 

(Co,  N2,  Oe) 

115,720 

1,961 

(Ni,  N2,  Oe) 

113,230 

1,936 

(Pb,  Ni,  Oe) 

105,400 

509 

98,200 

474 

(Cu,  Nt,  Oe) 





82,230 

1,295 

(Fe,  Ni,  0») 



...... 

157,150 

2,806 

(H,  N,  Oi)  (gas) 

34,400 

34,400 

48,800 

48,800 

(Hg,  N,  0») 



28,900 

145 

(Hg.  N2,  Oe 



...;.. 

54,000 

270 

(Ag,  N,  Oi) 

28,700 

275 

23,000 

213 

HEAT  OF  FORMATION  OF  NITRATES 
B.  From  Metallic  Oxide  and  -A/205 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

NiOs 

Nitrate 

Metallic 
Oxide 

N206 

Nitrate 

(K2O,  N2OB) 
(BaO,  N2Os) 
(Li2O,  N2Os) 
(Na2O,  N2O5) 
(SrO,  N2Ofi) 
(CaO,  N2O8) 
(MgO.  N2O5) 
(MnO,  N206) 
(ZnO,  N2Os) 
(T1«O,  N2O6) 
(FeO,  N2O6) 
(CoO,  N206) 
(NiO,  N2O8) 
(PbO,  N2O5) 
(CuO,  N20B) 
(Fe2O3,  3N2O6) 
(H«0.  N20s) 
(Hg2O,  N2O6) 
<HgO,  N2Os) 
(Ag2O,N2Os) 

141,000 
92,600 
86,220 
119,800 
87,400 
69,300 

1,500 
.605 
2.874 
1,932 
848 
1,238 

1,306 
857 
798 
1,109 
809 
642 

698 
354 
583 
705 
414 
423 

124,400 
83,200 
86,820 
110,400 
82,300 
73,300 
61,600 
54,800 
45,700 
52,360 
52,100 
50,420 
50,530 
46,200 
43,330 
115,100 
27,400 
34,400 
31,300 
37,800 

1,323 
544 
2,894 
1,781 
799 
1,309 
1,540 
772 
564 
124 
724 
672 
678 
207 
545 
719 
1,522 
827 
145 
163 

1,152 
770 
804 
1,022 
762 
679 
570 
507 
423 
485 
482 
467 
468 
428 
401 
1,066 
254 
319 
290 
350 

616 
319 
587 
649. 
390 
447 
416 
306 
242 
98 
274 
276 
277 
140 
231 
428 
214 
656 
97 
111 

72,300 

171 

669 

136 
161 

53,400 

240 

494 

-1,400 

-78 

-13 

-11 

49,200 

212 

456 

145 

APPLICATIONS  OF  THERMOCHEMISTRY. 


31 


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32 


METALLURGICAL  CALCULATIONS. 


HEAT  OF  FORMATION  OF  SULFATES 
A.  From  the  Elements 


Formula 

Anhydrous 

To  Dilute  Solution 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(Ki,  S,  00 

344,300 

4,414 

337,700 

4,340 

(Ba,  S,  OO 

339,400 

2,477 

(Liz,  S,  OO 

333,500 

23,820 

339,600 

24,260 

(Sr,  S,  OO 

330,200 

3,795 

(Na»,  S,  OO 

328,100 

7,133 

328,500 

7,141 

(Ca,S,  OO 

317,400 

7,935 

321,800 

8,045 

(Mg,  S,  OO 
(Ah,  Sa,  Ou) 
(N2,  H8,  S,  OO 

300,900 

12,540 

321,100 
879,700 
281,100 

13,380 
16,290 

7,808 

283,500 

7,875 

(Mn,  S,  OO 

249,400 

4,535 

263,200 

4,785 

(Zn,  S,  OO 

229,600 

3,532 

248,000 

3,815 

(Pe,  S,  00 





234,900 

4,195 

(Co,  S,  OO 



228,900 

3,880 

(Ni,  S,  OO 





228,700 

3,919 

(Fez,  Sa,  Oi2) 





650,500 

5,808 

(T12,  S,  00 

221,800 

544 

213,500 

523 

(Cd,  S,  OO 
(Pb,  S,  OO 
(H2,  S,  00 

219,900 
215,700 
192,200 

1,963 
1,042 
96,100 

231,600 

2.068 
105,100 

210,200 

(Cu,  S,  OO 

181,700 

2,861 

197,500 

3,110 

(Hg2,  S,  OO 

175,000 

438 





(Ag2,  S,  OO 

167,100 

774 

162,600 

753 

(Hg,  S,  00 

165,100 

826 

HEAT  OF  FORMATION  OF  SULFATES 
B.  From  Metallic  Oxide  and  S03 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

SOa 

Sulfate 

Metallic 
Oxide 

SOa 

Sulfate 

(KsO,  SO3) 

154.200 

1,640 

1,928 

886 

147,600 

1,570 

1,845 

848 

(BaO,  SOa) 

114,100 

745 

1,426 

490 



(Li2O,  SOa) 

105,800 

3,527 

1,322 

962 

111.900 

3,730 

1,399 

1,017 

(SrO,  SOa) 

107,100 

1,040 

1,339 

585 

(Na2O,  SOa) 

135,800 

2,190 

1,698 

956 

136.200 

2.197 

1,702 

959 

(CaO,  SOa) 

94,000 

1,679 

1,175 

692 

98,400 

1,757 

1,230 

723 

(MgO,  SOa) 

65,600 

1,640 

820 

547 

85,800 

2,145 

1,072 

715 

(AhOa,  3SOa) 



211,400 

2,072 

881 

618 

(MnO,  SOa) 

66,600 

938 

832 

441 

80,400 

1,132 

1,005 

532 

(ZnO,  SOa) 

52,900 

653 

661 

328 

71,300 

880 

891 

443 

(FeO,  SOa) 

77,300 

1,074 

966 

509 

(CoO,  SOa) 

72,900 

972 

911 

470 

(NiO,  SOa) 

75,300 

1,011 

941 

487 

(FeaOa,  3SOa) 

179,200 

1,120 

747 

448 

(ThO,  SOa) 

87,100 

205 

1,089 

173 

78,800 

186 

985 

156 

(CdO,  SOa) 

61,700 

482 

759 

297 

73,400 

573 

918 

353 

(PbO,  SOa) 

73.000 

328 

912 

241 

(H2O,  SOa) 

31,300 

1,739 

391 

319 

49,300 

2,739 

616 

503 

(CuO,  SOa) 

52,100 

655 

651 

326 

67,900 

853 

849 

425 

(Hg20,  S0,i 

60,900 

146 

761 

123 

(Ag2O,  SO.) 

68,200 

294 

852 

219 

63,700 

275 

796 

204 

(HgO,  SOt) 

51,700 

239 

646 

175 

APPLICATIONS  OF  THERMOCHEMISTRY. 
HEAT  OF  FORMATION  OF  BI-SULFATES 


33 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Bi-sulfate 

Metal 

Bi-sulfate 

(K,  H,  S,  OO 
(Na,  H,  S,  00 
(N,  H«,  S,  00 

276,100 
269.100 
244.600 

7,080 
11,700 
13.588 

2,030 
2.243 
2.127 

272,900 
268,300 
245,100 

6,974 
11.665 
13,617 

2,006 
2,236 
2,131 

HEAT  OF  FORMATION  OF  PHOSPHATES 
Anhydrous 


A.  From  the  Elements 


B.  From  Metallic  Oxide  and  PiOs 


Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit 

Weight 
of  Metal 

Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 

Phos- 

Oxide 

fifJt 

phate 

(Bas,  P2,  Os) 

951,700 

2,316 

(3BaO,  P2Os) 

186,200 

406 

1,311 

310 

(Sra,  Pi,  Os) 

945,000 

7.130 

(3SrO,  PiOs) 

186.100 

602 

1,311 

413 

(Ca3,  P2,  Os) 

919.200 

7,660 

(3CaO,  PjO.) 

159,400 

949 

1,123 

514 

(Mgs.  P,  O8) 

910,600 

12,647 

(3MgO,  PzOs) 

115,100 

959 

811 

439 

(Nai.  P,  OO 

452,400 

6,557 

(3NajO,  P«O.) 

238.300 

1,281 

1,678 

727 

(Mna,  Pt,  Os) 

737.500 

4,470 

(3MnO,  PzO6) 

99,500 

467 

701 

280 

HEAT  OF  FORMATION  OF  ARSENITES  AND  ARSENATES 
A.  From  the  Elements 


Anhydrous 


Formula  and 
Reaction 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(Na,  As,  O2) 

158,050 

6  872 

(Nas,  As,  O4) 
(Ks,  As,  O4) 

360,800 

5,229 

381,500 
396,200 

5,529 
3,386 

(Sr3,  As2,  Os) 
(Cas,  As2,  Os) 
(Mga,  As2,  Os) 
(Baj,  As2,  O«) 

761.000 
732,800 
712,600 
629,200 

2.915 
6,107 
9,897 
1,531 

To  Dilute  Solution 


HEAT  OF  FORMATION  OF  TUNGSTATES 
A  nhydrous 


A.  From  the  Elements 


B.  From  Metallic  Oxide  and  WOj 


Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit 
Weight 
of  Metal 

Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

WOt 

Tung- 
state 

(Na2,  W,  O4) 

391,400 

8.509 

(Na2O,  WOa) 

94,700 

1,527 

408 

322 

34 


METALLURGICAL  CALCULATIONS. 

B.  From  Metallic  Oxide  and  As203  or  As20t 


Anhydrous 

To  Dilute  Solution 

Formula 
and 
Reaction 

Per 
For- 

Per Unit  Weight  of 

Per 

For- 

Per Unit  Weight  of 

mula 
Weight 

Metallic 
Oxide 

AszOz  or 
AszOs 

Com- 
pound 

mula 
Weight 

Metallic 
Oxide 

AszOs  or 
As2O& 

Com- 
pound 

(Na2O,  As2O3) 

59  300 

9  565 

300 

227 

(3Na2O,  As20s) 

201,000 

1,081 

874 

483 

242,400 

1,303 

1,054 

583 

(3K2O,  As2O5> 

278  400 

987 

1  210 

K44. 

(3SrO,  As2Os) 

148,000 

567 

643 

302 

(3CaO,  As2O6) 

118,900 

708 

502 

299 

(3MgO,  As2Os) 

63,000 

525 

274 

151 

(3Ba2O,  As2O5) 

9,600 

21 

42 

14 

HEAT  OF  FORMATION  OF  BORATES.    A.  From  the  Elements 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

Per  Formula 
Weight 

Per  Unit  Weight 
of  Metal 

(Na»,  B4,  OT) 

748,100 

16,263 

758,300 

16,485 

B.  From  Metallic  Oxide  and  B203 


Formula 
and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 
For- 
mula 
Weight 

Per  Unit  Weight  of 

Per 
For- 
mula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

B20a 

Borale 

Metallic 
Oxide 

BzOt 

Borate 

(Na2O,  2B2Oa) 

102,500 

1,653 

732 

507 

112,700 

1,818 

805 

558 

HEAT  OF  FORMATION  OF  MOLYBDATES.    Anhydrous 


A.  From  the  Elements 

B.  From  Metallic  Oxide  and  MoOa 

Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit 
Weight 
of  Metal 

Formula    and 
Reaction 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

MoOs 

Molyb- 
date 

(Naz,  Mo,  04) 

363,800 

7,909 

(Na2O,  MoOa) 

88,400 

1,426 

612 

429 

HEAT  OF  FORMATION  OF  TITANATES.     Anhydrous 


A.  From  the  Elements 

B.  From  Metallic  Oxide  and  TiO2 

Formula  and 
Reaction 

Per 
Formula 
Weight 

Per  Unit 
Weight 
of  Metal 

Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

Ti02 

Titan- 
ate 

(Naz,  Ti,  Os) 

388,500 

8,446 

(Na2O,  TiO2) 

69,700 

1,124 

871 

491 

APPLICATIONS  OF  THERMOCHEMISTRY. 


35 


HEAT  OF  FORMATION  OF  MANGANATES  AND  PER-MANGANATES 
Anhydrous 


A.  From  the. Elements 


B.  From  Metallic  Oxide  and  Mn2O7 


Formula  and 
Reaction 

Per 
Formula 
Weight 

Per  Unit 
Weight 
of  Metal 

Formula  and 
Reaction 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

MniQi 
MniOi 

Com- 
pound 

(Nai,  Mn,  O4) 

269,400 

5,857 

(NazO.  Mn03) 

(K,  Mn,  04) 

200,050 

5,129 

(K20,  MnzOy) 

HEAT  OF  FORMATION  OF  ALUMINATES 
A  nhydrous 


A.  From  the  Elements 


B.  From  Metallic  Oxide  and  AhOa 


Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit 
Weight 
of  Metal 

Formula  and 
Reaction 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Metallic 
Oxide 

AhOi 

Alumin- 
ate 

(Na,  Al,  O2) 

533,000 

11,587 

(Na2O,  A1203) 

40,000 

645 

394 

244 

(Ca,  Ah,  04) 

524,550 

13,114 

(CaO,  A12O3) 

450 

8 

4 

3 

(Ca2,  Ah.  Os) 

658,900 

8,236 

(2CaO,  AhOO 

3,300 

29 

32 

15 

(Caa,  Ah,  Oe) 

780,050 

6,500 

(3CaO,  AhOi) 

-7,050 

-42 

-69 

-26 

HEAT  OF  FORMATION  OF  CYANIDES 


Anhydrous 


Formula  and 
Reaction 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Metal 

CN 

Cyanide 

Metal 

CN 

Cyan- 
ide 

(Ca,  C*.  N2) 

41,650 

1,041 

801 

453 

(K,  C,  N) 

33,450 

858 

1,287 

515 

30,250 

776 

1,163 

465 

(Na.  C,  N) 

25,950 

1,128 

998 

530 

25,450 

1,107 

979 

519 

(Sr,  C2,  N2) 

47,000 

540 

904 

338 

(Ba,  C2)  NO 

48,300 

353 

929 

256 

(Ca,  Ct,  NO 

41,650 

1,041 

801 

453 

(Mg,  C2,  N2) 

34,000 

1,417 

654 

447 

(Ni,  C«.  Ni) 

-23,400 

-400 

-450 

-212 

(Zn,  C2,  NO 

-24,550 

-378 

-472 

-210 

(Per,  Cis,  NiO 

-256,700 

-655 

-549 

-298 

(Cd,  C2,  NO 

-31,850 

-284 

-612 

-194 

(Cu,  C,  N) 

-20,375 

-320 

-784 

-227 

(Pd,  Cx  N») 

-49,250 

-465 

-947 

-312 

(gas) 

(H,  C,  N) 

-27,150 

-27,150 

-1,044 

-1,006 

-21,050 

-21,050 

-810 

-781 

(Hg,  C2,  NO 

-59,150 

-296 

-1,137 

-235 

(Ag,  C,  N) 

-34,000 

-315 

-1,308 

-254 

To  Dilute  Solution 


36 


METALLURGICAL  CALCULATIONS. 


HEAT  OF  FORMATION  OF  CYANATES 
A.  From  the  Elements 


Formula  and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Metal 

Cyanate 

Metal 

Cyanate 

(K,  C.  N.  0) 

105,850 

2,714 

1,307 

100,650 

2,581 

1,242 

(Na,  C,  N,  0) 

105,050 

4,567 

1,616 

100,250 

4,359 

1,542 

(Ag,  C,  N,  O) 

26,450 

245 

176 

(Hg,  Ci.  Ni,  Oi) 

-62,900 

-315 

-221 

B.  From  Cyanide  and  Oxygen 


Anhydrous 

To  Dilute  Solution 

Formula  and 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Reaction 

Formula 

Formula 

Weight 

Cyan- 
ide 

Oxygen 

Cyanate 

Weight 

Cyan- 
tde 

Oxygen 

Cyanate 

(KCN,  0) 

72,400 

1,115 

4,525 

894 

67,200 

1,034 

4,200 

830 

(NaCN,  O) 

79,100 

1,614 

4,944 

1,217 

74,300 

1,516 

4.644 

1,143 

(AgCN.  0) 

60,450 

451 

3,781 

403 

(HgCjN,,  02) 

-3,750 

-15 

-117 

-13 

HEAT  OF  FORMATION  OF  METALLO-CYANIDES 
A.  From  the  Elements 


Anhydrous 

To  Dilute  Solution 

Formula  and 
Reaction 

Per 

Per  Unit  Weight  of 

Per 

Per  Unit  Weight  of 

Weight 

Metal 

CN 

Cyanide 

Weight 

Metal 

CN 

Cyanide 

(K4.  Pe,  Ce,  Ne) 

157,300 

1,007 

1,007 

428 

145,300 

931 

931 

395 

(H4,  Fe,  Ce.  Ne) 

-  102,000 

-25,500 

-652 

-472 

-101,500 

-25,375 

-650 

-470 

(Ki,  Pe,  Ce,  Ne) 

129,600 

1,108 

831 

394 

100,800 

862 

647 

307 

(Hi.  Pe,  Ce,  Ne) 

-127,400 

-42,467 

817 

593 

(K.'Ag,  C«.  Ni) 

13,700 

351 

264 

69 

5,350 

137 

103 

27 

B.  From  the  Constituent  Cyanides 


Formula  and 
Reaction 

Anhydrous 

To  Dilute  Solution 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Per 
Formula 
Weight 

Per  Unit  Weight  of 

Cyan- 
ide I 

Cyan- 
ide II 

Com- 
pound 

Cyan- 
ide I 

Cyan- 
ide II 

Com- 
pound 

(KCN.  AgCN) 

-86,100 

-1.325 

-642 

-433 

-94.450 

-  1,452 

-704 

-474 

APPLICATIONS  OF  THERMOCHEMISTRY. 


37 


HEAT  OF  FORMATION  OF  AMALGAMS 


Formula  and 
Reaction 

To  Formula  Given 

Per 

Formula 
Weight 

Per  Unit  Weight  of 

Mercury 

Metal                   Amalgam 

(Hg,  Na) 
(Hg2,  Na) 
(Hg4,  Na) 
(Hg«,  Na) 
(Hg,,  Na) 
(Hg2,  K) 
(Hg4t  K) 
(Hg9,  K) 
(Hg12,  K) 
(Hg,,  K) 
(Hg,,  Ag) 
(Hg,,  Au) 
(Hg,.«,  Zn)     \ 
8.7%  Zn           f 
(Hgo.94,  Zn      \ 
25.9%  Zn        / 

10,300 
17,800 
20,000 
21,900 
19,000 
20,000 
29,700 
33,000 
34,600 
25,600 
2,470 
2,580 

-1,345 
-703 

52 
45 
25 
18 

448 
774 

870 
952 
826 
513 
762 
846 
887 
656 
23 
13 

-21 
-11 

46 
42 
24 
18 

50 

37 
18 
14 

46 
35 
18 
14 

-2 

-3.8 

-1.8 

-2.8 

HEAT  OF  FORMATION  OF  ALLOYS 


Formula 
or 
Composition 

To  Formula  or  Composition  Given 

Percentage  Composition 

Per 
Formula 
Given 

Per  Unit  Weight  of 

Metal  I 

Metal  11 

Alloy 

Metal  I 

Metal  II 

Per  Cent. 

Per  Cent. 

(Cu,  Znj) 

10,143 

160 

78 

52 

Cu        32.8 

Zn        67.2 

(Cu,  Zn) 

5,783 

91 

89 

45 

Cu        49.4 

Zn        50.6 

(Cus,  Al) 

26,910 

141 

997 

124 

Cu       87.6 

Al         12.4 

(Cu2.  Al) 

21,278 

167 

788 

138 

Cu       82.5 

Al         17.5 

(Cua,  Ah) 

17,395 

91 

322 

71 

Cu       77.9 

Al         22  .  1 

(Cu.  Al) 

1,887 

30 

70 

21 

Cu        70.2 

Al         29.8 

(Cu2,  All) 

10,196 

80 

126 

49 

Cu       61.1 

Al         38.9 

(Cu,  Ah) 

31,900 

502 

591 

271 

Cu       54.0 

Al         46.0 

(Nat.  K)  (liquid) 

-2,930 

-64 

-75 

-34 

Na       54.1 

K         45.9 

(Na,  K2)  (liquid) 

1,940 

84 

25 

19 

Na       22.77 

K         77.23 

(Na,  Ka)  (liquid) 

1,160 

50 

10 

8 

Na       16.43 

K         83.57 

(Pbu,  Zn) 

23,188 

6 

357 

5.8 

Pb        98.4 

Zn          1.6 

(Pb,  Zn) 

952 

4.6 

15 

+  3.5 

Pb        76.1 

Zn        23.9 

(Pbi.2«,  Bi) 

-  1,782 

-6.9 

-8.6 

-3.8 

Pb        55.6 

Bi         44.4 

(Sne.i,  Zn) 

-4,789 

-6.6 

-74 

-6.1 

Sn        91.7 

Zn          8.3 

(Pb,  Snu.r) 

-8,034 

-39 

-4.3 

-3.9 

Pb        10.0 

Sn        90.0 

(Pb,  Sm.s) 

-1,028 

-5.0 

-3.1 

-1.9 

Pb        38.2 

Sn        61.8 

(Pbi.i,  Sn) 

450 

1.0 

3.5 

0.8 

Pb        79.0 

Sn        21.0 

(Pbio.8,  Sn) 

2,594 

1.2 

22 

1.1 

Pb        95.0 

Sn          5.0 

(Pbt«.  Sn) 

5,914 

1.0 

50 

1.0 

Pb        98.0 

Sn          2.0 

(Mg,  Zm) 

24,900 

1,038 

192 

162 

Mg       15.6 

Zn        84.4 

(Mg4,  Ah) 

164,800 

1,717 

204 

93 

Mg       54.2 

Al         45.8 

(Mg,  Cd) 

17,700 

738 

158 

130 

Mg       17.6 

Cd       82.4 

(Na,  Cd.) 

30,800 

1,339 

137 

125 

Na         9.3 

Cd       90.7 

(Na,  Cd«) 

60,600 

2,635 

108 

104 

Na          3.9 

Cd       96.1 

(Ca,  Zm) 

55.600 

1,390 

214 

185 

Ca        13.3 

Zn        86.7 

(Ca,  Zmo) 

199,100 

4,978 

306 

29 

Ca          5.8 

Zn        94.2 

(Cui,  Cdi) 

47,700 

376 

142 

103 

Cu        27.4 

Cd        72.6 

38 


METALLURGICAL  CALCULATIONS. 


THERMOCHEMICAL  CONSTANTS  OF  BASES  TO  DILUTE  SOLUTION. 


(Per  Chemical  Equivalent  of  Base.} 
'For  explanation  of  nature  and  use  of  these  tables,  see  page  16. 


Caesium 

Rubidium 

Lithium 

Potassium 

Beryllium  (Glucinum) 

Barium 

Thorium 

Strontium 

Sodium 

Lanthanum 

Neodymium 

Calcium 

Praesodymium 

Magnesium 

Aluminium 

Ammonium  (from  the  elements) 

Vanadium 

Cerium 

Titanium 

Uranium 

Vanadium 

Vanadium 

Silicon 

Boron 

Vanadium 

Zirconium 

Manganese 

Zinc 

Chromium 

Phosphorus 

Lithium  (per-salts) 

Barium  (per-salts) 

Iron  (ferrous) 

Tungsten  (Wolfram) 

Tungsten  (Wolfram) 

Cadmium 

Cobalt      . 

Manganese 

Sodium  (per-salts) 

Nickel 

Antimony 

Arsenic 

Sulfur 


Cs+ 
Rb+ 
Li+ 
K+ 


Na 


(N,  H4) 


+81,240 
76,265 
62,900 
61,900 
60,350 
59,950 

58,700 
57,200 

55,600 
54,400 

54,300 
40,100 
33,400 


24,900 
17,200 


10,900 


9,000 
8,200 


7,700 
2,800 


APPLICATIONS  OF  THERMOCHEMISTRY. 


39 


Iron  (ferric) 

Tin  (stannous) 

Tin  (stannic) 

Lead 

Carbon 

Chromium 

Bismuth 

Antimony 

Arsenic 

Copper  (cuprous) 

Hydrogen 

Tellurium 

Tellurium 

Sulfur 

Selenium 

Selenium 

Tellurium 

Copper  (cupric) 

Lead  (per-salts) 

Thallium 

Carbon 

Hydrogen  (per-salts) 

Mercury  (mercurous) 

Palladium 

Mercury  (mercuric) 

Platinum 

Silver 

Gold  (auric) 

Gold  (aurous) 


27,30 
1,900 


Cu+ 

H+ 

Te+ 


Ag 
Au 
Au 


400 


±0 
-900 

-1,300 


-7,900 


•14,250 
19,450 
25,300 

-30,300 


THERMOCHEMICAL  CONSTANTS  OF  ACIDS  TO  DILUTE  SOLUTION 
(Per  Chemical  Equivalent  of  Acid  Element  or  Radical] 


Fluorine 

Chlorine 

Bromine 

Oxygen 

lodine 

Sulfur 

Selenium 

Tellurium 


Hydrate 
Sulf  -hydrate 


Elements 


F~ 

Cl~ 

Br~ 


I" 


Acid  Radicals  (from  the  elements) 

(O,  H)~ 
(S,  H)~ 


52,900 

39,400 

27,500 

20,700 

13,200 

-5,100 

-17,900 


55,200 
3,400 


40 


METALLURGICAL  CALCULATIONS. 


Selen-hydrate 

Hypochlorite 

Chlorate 

Per-chlorate 

Hypobromite 

Bromate 

lodate 

Per-iodate 

Hypo-sulfite 

Sulfite 

Bi-sulfite 

Pyro-sulfite 

Sulfate 

Bi-sulfate 

Per-sulfate 

Di-thionate 

Tri-thionate 

Tetra-thionate 

Penta-thionate 

Selenite 

Selenate 

Hypo-nitrite 

Nitrite 

Nitrate 

Phosphate 

Mono-H- Phosphate 

Di-H-Phosphate 

Arsenite 

Arsenate 

Mono-H- Arsenate 

Di-H-Arsenate 

Cyanide 

Cyanate 

Sulfo-cyanate 

Ferro-cyanide 

Fern-cyanide 

Carbonate 

Bi-carbonate 

Formate 

Acetate 

Oxalate 


(Se,  H)- 

19,100 

(Cl,  CO- 

27,500 

CCI,  oo- 

21,900 

(Cl,  04)- 

39,400 

(Br,  0)- 

22,800 

(Br,  03)- 

6,700 

(i,  oo- 

60,470 

(i,  oo- 

48,070 

^(S2,  03)~ 

71,750 

M(S,  03)~ 

75,100 

(H,  S,  03)- 

149,400 

K(S2,  oo— 

115,200 

^(S,  04)~ 

107,000 

(H,  S,  04)- 

211,100 

H(S2,  00" 

158,100 

^(S2,  0.)— 

138,350 

H(S>f  o.)— 

136,500 

K(S4,  06)~ 

130,600 

H(Si,  o«)— 

133,100 

H(Se,  0,)— 

60,050 

H(Se,  04)~ 

72,800 

(N,  0)- 

-3,800 

(N,  02)- 

27,000 

(N,  0,)- 

48,800 

M(P,  04)  

99,300 

H(H,  P,  04)" 

152,750 

(H2,  P,  04)- 

307,700 

H(As2,  04)~ 

102,150 

H(As,  04)  

70,200 

H(H,  As,  00— 

108,050 

(H2,  As,  04)- 

217,200 

(C,  N)- 

-34,900 

(C,  N,  0)-  ' 

37,100 

(C,  N,  S)- 

-18.100 

K(Pe,  C.,  N6)  

-25,600 

H(Fe,  C6,  NO  

-52,800 

H(C,  0,)— 

82,450 

(H,  C,  03)- 

169,100 

(C,  H,  00- 

104,600 

(C2,  H,,  02)- 

120,500 

H(Cfc  00" 

99,800 

CHAPTER  III. 
THE  USE  OF  THE  THERMOCHEMICAL  DATA. 

SIMPLE  COMBINATIONS. 

If  the  problem  is  the  simple  calculation  of  how  much  heat 
is  evolved  in  the  combination  of  a  given  weight  of  one  element 
with  another,  the  factors  needed,  the  heat  evolved  per  unit 
weight  of  substance  combining,  are  obtained  by  a  simple  divi- 
ion  from  the  thermochemical  data  given.  Thus,  suppose  the 
question  to  be  the  total  heat  evolved  in  Problem  3,  in  the 
oxidation  in  a  Bessemer  converter  of 

100  kilos,  of  carbon  to  carbonic  oxide. 

200  kilos,  of  carbon  to  carbonous  oxide. 
50  kilos,  of  manganese  to  MnO. 

150  kilos,  of  silicon  to  SiO2. 

500  kilos,  of  iron  to  FeO. 
The  solution  would  be 

Q7  200 
100X      is      =  100X8100  =      810,000  Calories. 

200  X  ??ii§2  =  200X2430  =     486,000 

LZ 

on  QDO 

50  X   J*   -  50X1653=   82,650 
oo 

150X18o'°OQ  =  150X7000  =  1,050,000 


500  X      'ft      =  500X1173  =      586,500 
Total  3,015,150 

COMPLEX  COMBINATIONS. 

If  the  problem  is  a  step  more  complex,  that  is,  includes  the 
combination  of  compounds  with  each  other  to  form  a  more 

41 


42  METALLURGICAL  CALCULATIONS. 

complex  compound,  the  molecular  weights  are  still  our  guide, 
together  with  the  thermochemical  data  given.  If,  for  in- 
stance, the  question  above  solved  is  complicated  by  the  further 
requirement,  to  add  the  heat  evolved  by  the  formation  of  the 
slag,  that  is,  of  the  MnO  and  FeO  with  SiO2  to  form  silicate, 
we  may  calculate  the  heat  of  union  of  FeO  and  MnO  with 
SiO2  as  follows: 

(Mn,  Si,  O3)  =  276,300  Calories. 
But  (Mn,  O)  =     90,900 

and  (Si,  O2)  =  180,000 

therefore  (MnO,  SiO2)  =      5,400 

or,  per  kilo,  of  MnO  =  -  :  -  =  76  Calories. 

71 

Similarly 

(Fe,  Si,  O3)  =  254,600  Calories. 

(Fe,  O)  =    65,700       " 

(Si,  O2)  =  180,000       " 

(FeO,  SiO2)  =      8,900 


per  kilo,  of  FeO  =  ~  =  124  Calories. 

{& 

Therefore  the  additional  heat  of  formation  of  the  slag  may  be 

Wt.  MnO  =     64.5  kilos.  X   76  =     4,902  Calories. 
Wt.  FeO    =  642.8      "     X124  =  79,707 

Sum  =  84,609 

Similar  principles  of  calculation  apply  to  all  the  oxygen- 
containing  salts.  Thus,  if  from  the  heat  of  formation  of  any 
sulphate  we  subtract  the  heat  of  formation  of  SO3,  and  also  of 
the  metallic  oxide  present,  the  residue  is  the  heat  of  combina- 
tion of  the  metallic  oxide  with  SO3,  —  the  weights  involved 
being  always  those  represented  by  the  formulae.  Thus,  calling 
MO  any  metallic  oxide,  we  may  express  the  principle  as  follows: 

(MO,  SiO2)  =  (M,  Si,  O3)   —  (M,  O)—  (Si,  O2) 

(MO,  SO3)  =  (M,  S,  O3)     —  (M,  O)—  (S,  O3) 

(MO,  CO2)  =  (M,  C,  O3)    —  (M,  0)—  (C,  O2) 

(3MO,  P2O5)  =  (M3,  P2,  O8)  —  3(M,  0)—  (P2,  O5) 
etc,  etc. 


THERMOCHEM1CA  L  DATA.  43 

Example. — What  is  the  heat  required  to  calcine  limestone? 

(CaO,  CO2)  =  (Ca,  C,  O3)— (Ca,  O)  — (C,  O2) 
=     273,850  —131,500—97,200 
45,150  Calories. 

This  is  the  heat  required  to  split  up  CaCO3  (100  parts)  into 
CaO  (56)  and  CO2  (44) ;  the  heat  required    is  therefore 

45,150-i- 100  =    451.5  Calories  per  kilo,  of  CaCO3  decomposed. 
45,150-^44=1026.          "          "       "         CO2  driven  off. 
45,150^-56  =    806.          "          "       "         CaO  remaining. 

And  either  of  these  quantities  may  be.  used,  according  to 
convenience  in  working  the  problem. 

DOUBLE  DECOMPOSITIONS. 

If  the  thermochemical  problem  involves  the  simultaneous 
decomposition  of  one  or  more  substances  and  formation  of  one 
or  more  others,  then  the  chemical  equation  should  be  written, 
and  a  thermochemical  interpretation  given  of  all  the  energy 
involved  in  the  passage  from  starting  compounds  to  products. 
Every  chemical  equation  can  be  thus  interpreted,  if  the  heats 
of  formation  of  all  the  compounds  represented  in  it  are  known. 
We  obtain  the  net  energy  of  the  reaction  by  assuming  that  all 
the  substances  used  are  resolved  into  their  elements,  and  that 
all  the  products  are  formed  from  their  elements;  the  first  item 
is  therefore  to  add  together  the  heats  of  formation  of  all  the 
substances  used,  starting  with  their  elements,  and  by  changing 
the  algebraic  sign  of  this  sum  we  have  the  heat  necessary  to 
decompose  the  substances  used  into  their  elements;  the  second 
item  is  similarly  found  by  adding  together  the  heats  of  forma- 
tion of  all  the  substances  formed;  the  difference  between  these 
two  items  is  the  net  energy  of  the  reaction.  In  making  these 
summations,  regard  must,  of  course,  be  paid  to  the  number  of 
molecules  of  each  substance  concerned,  as  shown  in  the  re- 
action (because  the  tabulated  data  are  the  heats  of  formation 
of  one  molecule  only)  and  to  the  algebraic  sign  of  the  heats  of 
formation. 

Examples. — (a)  What  heat  is  evolved  when  dry  ferric  oxide 
is  reduced  by  aluminium,  in  the  Goldschmidt  "  Thermite  " 
process? 


44  METALLURGICAL  CALCULATIONS. 


Fe2O3  +  2Al  =      Al2O3  +  2Fe 


—195,600 


+  392,600 


+  197,000  Calories. 

(b)  Write   the   equation   showing  the   reaction   of  metallic 
sodium  on  water  in  excess: 

2H20  +  2Na  =  2NaOH  +  H2  +  aq.  (excess  water) 


—2(69,000) 


—138,000 


+  2(112,500) 


+  225,000 


+  87,000  Calories. 

(c)  What  heat  is  evolved  in  the  reaction  of  water*  on    cal- 
cium carbide  to  form  acetylene  gas? 

CaC2     +2H2O  =  Ca02H2+     C2H2 


—  (—6250)— 138,000 


—131,750 


+  215,600 +(—54,750) 


+  160,850 


+  29,100 

The  last  case  leaves  out  the  very  small  heat  of  solution  of 
the  calcium  hydrate  which  would  normally  go  into  solution. 
If  such  a  large  excess  of  water  was  used  that  all  the  calcium 
hydrate  formed  could  dissolve,  the  heat  of  formation  of  the 
hydrate  in  dilute  solution,  219,500  Calories,  would  be  used,  and 
the  final  result  would  be  33,000  Calories.  The  case  is  also  very 
instructive,  because  it  contains  two  compounds  which  are 
endothermic,  that  is,  their  heat  of  formation  is  negative,  and 
therefore,  as  in  the  case  of  CaC2,  heat  is  given  out  when  it 
decomposes,  while  in  the  case  of  C2H2  heat  is  absorbed  when  it 
is  formed. 

(d)  What  are  the  heats  of  combustion  of  the  gaseous  hydro- 
carbons which  form  constitutents  of  common  fuels,  expressed 
per  cubic  meter  of  gas  burning,  and  with  the  water  formed 
assumed  remaining  as  vapor,  uncondensed?  To  calculate  these 
we  write  the  equations  of  combustion,  with  water  as  gas,  and 
find  the  sum  total  of  heat  evolved;  since  every  molecule  of  gas 
burning  represents  22.22  m3  of  gas,  we  can,  by  a  simple  division, 
obtain  the  value  sought. 


THERMOCHEMICAL  DATA.  45 

Methane: 

CO2  +  2H2O 
+  97,200  +  2(58,060) 


—22,250 


+  191,070  Calories.  =  8,598  per  m8  CH«. 

Ethane: 

2C2H6  +  702  =         4CO2  +  6H2O 
—2(26,650)     |  +4(97,200)  +  6(58,060) 

+  683,860  Calories.    =15,387    perm3C2H8. 
Propane: 

C3H8  +  502  =         3CO2  +  4H2O 


—  33,850 


+3(97,200)  +4(58,060) 


+  489,990  Calories.         =  22,050  per  m3  C3H8. 

By  applying  these  principles  to  all  the  hydrocarbons  whose 
heat  of  formation  has  been  given  in  the  tables,  we  obtain  the 
following  useful  table,  water  being  considered  as  uncondensed 
and  cold: 

Molecular  Heat  of 

Combustion  .  Heat  of  Combustion  of 

Hydrocarbon.  Calories.    1  m3  (Calories)    1  ft3  (B.T.U.) 

Methane  (gas)  ..........  191,070  8,598  966 

Ethane  (gas)  ...........  341  ,930  15,387  1728 

Propane  (gas)  ..........  489,990  22,050  2477 

Ethylene  (gas)  ..........  321  ,770  14,480  1627 

Propylene  (gas)  .........  471  ,830  21,232  .2385 

Toluene  (liquid)  ........  906,990  ......  ____ 

B                (liquid  .......  758,130  ...... 

(gas  ..........  765,330  34,440  3869 

-  ,.        j  liquid  .....  1,428,930  ......  ____ 

Turpentine   |        ....... 


AT  (solid  .....  1,223,690 

Naphthaline   j 


Anthracine  (solid)  ......  1,690,150  ......  ____ 

Acetylene  (gas)  .........    307,210  13,825  1555 


TUT  xt,  1     1     LI  (liquid..     148,270 
Methyl  alcohol  |  ^  ^^ 


7,050  799 

_,,,  j  liquid...     295,330  

Ethyl  alcohol  j  gas ^^  13  7M  1M4 


46  METALLURGICAL  CALCULATIONS. 

Molecular  Heat  of 

Combustion.  Heat  of  Combustion  of 

.  (liquid 396,130  

(gas 403,630  18,163  2040 

(To  the  above  we  will  add  data  for  three  other  common 
gaseous  fuels.) 

Carbonous  oxide  (gas)             68,040  3,062  344 

Hydrogen  (gas) 58,060  2,613  293 . 5 

Hydrogen  sulphide 122,520  5,513  619 

CALORIFIC  POWER  OF  FUELS. 

By  using  the  principles  explained  we  can  calculate  the  cal- 
orific power  of  any  combustible,  with  the  aid  of  one  or  two 
simple  assumptions.  Regarding  the  water  formed,  we  will  con- 
sider in  all  ordinary  metallurgical  calculations  that  it  remains 
uncondensed,  thus  putting  it  on  the  same  footing  as  the  other 
products  of  combustion.  This  amounts  virtually  to  assuming 
that  the  latent  heat  of  condensation  of  the  vapor  formed  has 
not  been  generated  in  the  furnace,  an  assumption  which  is 
quite  justified  if,  as  is  almost  always  the  case,  the  water  formed 
inevitably  escapes  as  vapor.  If,  in  any  special  case,  the  pro- 
ducts of  combustion  are  in  reality  cooled  down  so  far  that  the 
water  does  condense,  then  it  would  be  proper  to  assume  the 
higher  calorific  powers  for  the  combustion  of  the  hydrogen- 
containing  materials,  and  thus  a  more  accurate  heat-balance  of 
the  furnace  operation  could  be  constructed.  An  instance  of 
the  latter  might  be  the  case  of  the  partial  combustion  of  a  fuel 
in  a  gas  producer,  where  the  fuel  gas  is  afterwards  cooled  be- 
fore using,  in  order  to  condense  from  it  ammonia  water,  etc. 
(Mond's  producer).  In  this  case,  it  is  possible  that  some 
water  formed  by  the  partial  combustion  would  be  afterwards 
condensed,  and  to  obtain  a  correct  heat-balance-sheet  of  the 
producer  it  would  be  necessary  to  assume  that  this  heat  had 
also  been  generated.  Such  cases  occur  very  rarely  in  metal- 
lurgical practice,  and  it  is  therefore  recommended  to  always 
use  the  lowrer  calorific  values,  assuming  water  vapor  uncon- 
densed and  its  latent  heat  of  condensation  not  to  have  been 
generated — unless  the  conditions  of  the  problem  point  plainly 
to  the  opposite  course  as  correct.  It  is  no  more  correct  to 
charge  against  a  furnace  the  latent  heat  of  condensation  of  the 


THERMOCHEMICAL  DATA.  47 

water  vapor  formed,  than  it  would  be  to  charge  against  it  the 
latent  heat  of  condensation  of  the  carbonic  oxide  formed,  if  the 
conditions  of  practice  are  such  that  it  is  impossible  to  utilize 
in  any  useful  manner  either  one  of  them.  Those  who  persist 
in  assuming  that  the  latent  heat  of  vaporization  of  the  water 
formed  by  combustion  is  generated  in  and  chargeable  against 
the  furnace,  and  then,  of  necessity,  charge  the  same  quantity 
against  the  heat  carried  off  in  the  products  of  combustion,  are 
in  the  great  majority  of  cases  adding  to  both  sides  of  the  balance 
sheet  a  quantity  which  cannot  possibly  be  utilized,  and  are 
therefore  distorting  all  the  factors  of  heat  generation  and 
distribution.  By  thus  abnormally  increasing  the  sensible  heat 
in  the  waste  gases,  they  abnormally  decrease  the  proportionate 
value  of  other  items  of  heat  distribution  and  utilization.  I 
have  dwelt  on  this  matter  at  length,  because  it  is  of  importance 
in  furnaces  using  fuel,  such  as  gas,  rich  in  hydrogen;  where,  in 
some  cases,  the  counting  of  the  latent  heat  of  condensation  of 
the  water  vapor  formed  in  the  escaping  gases,  would  perhaps 
double  the  apparent  chimney  loss,  and  give  distorted  values 
to  the  whole  heat  balance  sheet.  What  we  are  after,  in  every 
case,  are  the  real  facts  and  figures  as  to  the  working  of  a  furnace 
or  operation,  and  we  must  therefore  judge  the  operation  by 
a  theoretically  perfect  standard,  it  is  true,  but  not  by  a  theo- 
retically impossible  standard;  i.e.,  our  standard  must  be  the 
theoretically  possible  one  under  the  practical  conditions  neces- 
sarily prevailing.  A  radical  reform  is  very  much  needed  in 
just  this  respect, — in  the  treatment  of  the  latent  heat  of  vapor- 
ization of  the  water  formed — by  writers  on  metallurgical  pro- 
cesses and  problems  of  combustion. 

Problem  4. 

A  natural  gas  from  Kokomo,  Ind.,  contained  by  analysis: 
CH4  94.16,  H2  1.42,  C2H4  0.30,  CO2  0.27,  CO  0.55,  O2  0.32,  N2 
2.80,  H2S  0.18  per  cent. 

Required : 

(1)  What  is  its  metallurgical  calorific  power  per  cubic  meter 
and  per  cubic  foot? 

(2)  Comparing  it  with  coal,  having  a  calorific  power  of  8,000 
Calories,  how  much  gas  gives  the  same  generation  of  heat  as 
a  metric  ton  of  coal? 


48  METALLURGICAL  CALCULATIONS. 

(3)  If  the  natural  gas  costs  $0.15  per  1,000  cubic  feet,  at 
what  price  per  metric  ton  would  the  coal  furnish  heating  power 
at  the  same  cost  for  fuel? 

Solution.  —  Requirement  (1)  Heat  of  combustion  of  1  cubic 
meter: 

CH40.9416X   8,598  =  8095.9  Calories. 

H20.0142X   2,613  =      37.1 

C2H4  0.0030  X  14,480  =      43.4 

CO  0.0055X   3,062  =       16.8 

H2S0.0018X   5,513  =        9.9 

Total  -  8203.1 
In  British  thermal  units  per  cubic  foot  we  have 

8,203.1X3.967     ^   35.314  = 
8,203.1X0.11233  =  921.5     B.  T.  u.  per  ft3  (1) 

Requirement  (2): 

8000X1000 


8,203.1 

=  34,450  ft3. 

If  the  ton  of  coal  be  taken  as  2240  pounds,  instead  of  the 
metric  ton  of  2204  pounds,  the  equivalent  volume  is 


|r±r  =  35j013  ftsB 

Requirement  (3): 

34  450 

T6oo~x0'15  =  $5-16*  Per  metric  ton- 

or,  per  ton  of  2240  pounds, 

2240 
S5.16JX  2204  =  $5'25  per  long  ton. 

DULONG'S  FORMULA. 

When  a  solid  or  liquid  carbonaceous  fuel  is  to  be  burned, 
its  calorific  power  may  either  be  determined  directly  in  a  calori- 
meter (which  is  the  best  way,  if  carefully  done),  or  may  be 
calculated  from  its  analysis.  In  case  it  is  determined  calori- 


THERMOCHEMICAL  DATA.  49 

metrically,  the  weight  of  water  produced  per  unit  of  fuel  should 
be  determined  either  in  the  same  or  by  a  separate  experi- 
ment, and  the  latent  heat  of  vaporization  of  this  water  sub- 
tracted from  the  calorimetric  value  of  the  fuel,  in  order  to  get 
its  metallurgical,  or  practical,  calorific  power. 

Example. — A  coal  gave  9215  calories  per  gram  in  the  calori- 
meter, and  its  products  of  combustion  gave  up  0.45  grams  of 
condensed  water.  What  is  its  practical  calorific  power? 

Taking  the  products  cold,  the  heat  of  condensation  per  gram 
of  water  vapor  may  be  taken  as  606.5  calories  (Regnault),  and 
we  therefore  have  obtained  in  the  calorimeter 

0.45X606.5  =  273  Calories 

more  than  if  the  products  were  cold  and  water  uncondensed. 
The  practical  calorific  power  is  therefore 

9215  -  273  =  8942  Calories  per  kilogram. 

Dulong's  simple  formula,  or  method  of  calculation,  is  the  state- 
ment that  the  calorific  power  of  a  fuel  can  be  calculated  from 
the  amount  of  carbon,  hydrogen,  and  oxygen  it  contains  by 
assuming  all  the  carbon  free  to  burn,  the  oxygen  to  be  all 
combined  with  hydrogen  in  the  proportions  of  O2  to  2H2  (32 
to  4),  and  that  the  rest  of  the  hydrogen  is  free  to  burn. 

Example. — Taking  the  bituminous  coal  of  Problem  1,  con- 
taining carbon  73.60,  hydrogen  5.30,  nitrogen  1.70,  sulphur  0.75, 
oxygen  10.00,  moisture  0.60,  ash  8.05,  its  calculated  calorific 
power  is  per  kilogram: 

Carbon        0.7360 X    8100  =  5962  Calories. 
Hydrogen   0.0530 
0.0125 

0.0405  X  34,500  =  1397        "       (water  liquid) 
Sulphur  0.0075  X     2164=      17 

Total    7376 

This  calculation  is  made,  however,  on  the  assumption  that 
all  the  water  in  the  products  is  condensed  to  liquid,  i.e.,  not 
only  the  water  formed  by  combustion  of  the  free  hydrogen, 
but  also  the  already-formed  H2O  containing  the  oxygen  of  the 
coal,  as  well  as  the  moisture  present  to  start  with.  To  obtain 


60  METALLURGICAL  CALCULATIONS. 

the  practical  calorific  power  by  calculation  we  must  subtract 
from  the  above-generated  heat  the  latent  heat  of  all  the  vapor 
of  water  in  the  products,  as  follows: 

Moisture  in  coal 0 . 0060  kilos. 

Moisture  formed  by  combined  hydrogen .  .  0 . 1 125      " 
Moisture  formed  by  free  hydrogen 0.3645      " 

Total. 0.4830     " 

Latent  heat  of  vaporization  =  606.5x0.4830  =  293  Calories. 

Practical  calorific  power,  water  all  as  vapor  =  7083  Calories. 

The  values  thus  calculated  are  found  to  agree  satisfactorily 
with  the  laboratory  and  the  practical  calorific  powers  of  the 
fuels. 

THE  THEORETICAL  TEMPERATURE  OF  COMBUSTION. 

If  we  start  with  a  cold  fuel  and  cold  air  the  maximum  tem- 
perature which  can  be  obtained  by  the  combustion  is  simply 
that  temperature  to  which  the  heat  generated  can  raise  the 
products  of  combustion,  assuming  that  all  that  heat  resides 
primarily  in  the  products  as  sensible  heat.  This  is  then  a  prob- 
lem in  physics,  in  which  we  have  a  known  amount  of  heat 
available,  and  the  question  is  to  what  temperature  it  will 
raise  the  products  of  combustion.  The  problem  is  at  once 
soluble  when  we  know  the  mean  specific  heats  of  the  products 
of  combustion  and  their  respective  quantities. 

Until  a  few  years  ago  these  specific  heats  at  high  tempera- 
tures were  unknown,  and  very  false  results  were  obtained  by 
assuming  constant  specific  heats  from  ordinary  temperatures 
up,  and  values  were  thus  calculated  which  were  known  to  be 
hundreds,  and  in  some  cases,  thousands,  of  degrees  too  high. 
The  most  satisfactory  values  for  the  specific  heats  of  the  fixed 
gases  and  water  vapor  and  carbonic  oxide  are  those  deter- 
mined by  Mallard  and  Le  Chatelier,  and  their  proper  use  has 
entirely  solved  the  question  of  theoretical  temperatures  of  com- 
bustion, and  removed  a  positively  disgraceful  discrepancy 
between  theory  and  practice.  The  specific  heats  of  these  gases 
increase  with  temperature,  so  that  the  actual  specific  heat  of  a 
cubic  meter  (measured  at  standard  conditions)  is,  at  the  tem- 
perature t, — 


THERMOCHEMICAL  DATA.  51 

for        N2,  O2,  H2,  CO  — S  =  0.303  +  0.0000541 
"      <  CO2— S  =  0.37   +0.00044t 

11     "          (vapor)  H2O— S  =  0.34   +  0.00030t 

For  calculating  the  quantity  of  heat  needed  to  raise  the  gas 
from  O°  to  t°,  however,  we  need  the  mean  specific  heats,  Sm, 
between  O°  and  t°.  These  are,  of  course,  the  above  constants 
plus  half  the  increase,  viz.: 

for  N2,  O2,  H2,  CO— Sm  -  0.303 +  0.000027t 

CO2— Sm  =  0.37   +  0.00022t 
(vapor)  H2O— Sm  =  0.34   +0.00015t 

and  the   quantity  of  heat  needed  to  raise   1   cubic  meter   (at 
standard  conditions)  from  O°  to  t°  is 

for  N2  O2  H2,  CO— Q  (o-t)  =  0.303t  +  0.000027t2 
CO2— Q  (o-t)  =  0.37t   +  0.00022t2 
H2O— O  (o-t)  =  0.34t   +  0.0001 5t2 

or,  finally,  the  quantity  of  heat  needed  to  raise  1  cubic  meter 
(measured  at  standard  conditions)  from  t  to  t'  is 

for  N2,  O2,  H2,  CO— Q  (t'-t)  = 

0.303  (t'-t) +0.000027  (t/2-t2) 
CO2— Q  (t'-t)  = 

0.37    (t  '-t)  +  0.00022    (t  /2-t2) 
H20— Q  (t'-t)  = 

0.34    (t'-t) +0.00015    (t'2-t2) 

And  the  mean  specific  heat,  Sm,  between  any  two  tempera- 
tures is 

for  N2  O2,  H2,  CO   — Sm  (t'-t)  = 

0.303  +  0.000027  (t'  +  t) 
CO2— Sm  (t'-t)  = 

0.37  +0.00022    (t'  +  t) 
H2O  — Sm  (t'-t)  = 

0.34  +0.00015    (t'  +  t) 
Examples : 

(1)  What  is  the  temperature  of  the  hottest  part  of  the  oxy- 
hydrogen  blowpipe  flame? 

According  to  the  equation  2H2  +  O2  =  2H2O,  the  hydrogen 
gas  forms  an  equal  volume  of  water  vapor.  (Equal  numbers 
of  molecules.)  The  heat  of  combustion  of  one  cubic  meter  of 
hydrogen  is  2613  Calories.  The  question  is,  therefore:  "To 


52  ME  1 A  LLURGIOA  L  CA  UCULA  TIONS. 

what  temperature  will  2613  Calories  raise  one  cubic  meter  of 
water  vapor?  "     The  answer  is,  using  the  data  above, 

Q  (0-t)   =  0.34t  + 0.00015t2  =  2613 
Whence  t  =  3191° 

(2)  What   is   the   maximum   temperature   of   the   hydrogen 
flame  burning  in  dry  air? 

The  heat  evolved  is,  as  before,  2613  Calories,  and  there  is 
formed  also  one  cubic  meter  of  water  vapor,  but  the  products 
will  contain  also  the  nitrogen  which  accompanied  the  0.5  cubic 

meter  of  oxygen  necessary  for  combustion,  viz.:  •     '     •  —  0.5 

U .  ^Uo 

=  1.9  cubic  meters,  and  this  is  heated  as  well  as  the  water 
vapor.     Therefore, 

Q   (0-t)  =  (0.34t  +  0.00015t2)  +  1.9  (0.303t  +  0.000027t2)  =  2613 
Whence  0.916t  +  0.0002013t2  =  2613 

And  t  =  2010° 

(3)  What  is  the  temperature  of  the  air-hydrogen  blowpipe, 
if  25  per  cent,  excess  of  air  is  used,  above  that  required  ? 

All  the  data  are  the  same  as  above,  except  that  to  the  pro- 
ducts will' be  added  0.25  (0.5  +  1.9)  =0.6  cubic  meters  of 
unused  air,  which  has  the  same  specific  heat  as  nitrogen,  and 
therefore  the  equation  becomes 

Q   (o-t)  =  (0.34t  +  0.00015t2)+2.5   (0.303t  +  0.000027t2)  =  2613 
Whence  t  =  1764° 

This  calculation  brings  out  very  clearly  the  uselessness  and 
ineffectiveness  of  using  more  air  than  is  theoretically  neces- 
sary; any  excess  simply  passes  unused  into  the  products  of 
combustion,  and  thus  reduces  their  maximum  possible  tem- 
perature. The  obtaining  of  the  maximum  possible  temperature 
depends  upon  accurately  proportioning  the  supply  of  oxygen 
or  air  to  the  quantity  of  gas  burned ;  an  excess  or  a  deficiency 
will  result  in  a  lower  temperature. 

COMBUSTION  WITH  HEATED  FUEL  OR  AIR. 

If  the  fuel  itself  or  the  air  which  burns  it  is  preheated,  the 
sensible  heat  in  either  one  or  in  both  is  simply  added  to  the 
heat  generated  by  the  combustion,  to  give  the  total  amount  of 


TEERMOCHEMICAL  DATA.  53 

heat  which  must  be  present  as  sensible  heat  in  the  products  of 
combustion.  The  effect  is  exactly  the  same  as  if  the  heat 
developed  by  combustion  had  been  increased  by  the  sensible 
heat  in  the  fuel  or  air  used. 

What  is  the  calorific  intensity  (theoretical  maximum  tem- 
perature) obtained  by  burning  carbonous  oxide  gas?  (a) 
Cold,  with  cold  air;  (b)  cold,  with  hot  air  at  700°  C.;  (c)  hot, 
with  hot  air,  both  at  700°  C. 

(a)  Take    one    cubic    meter   of    carbonous    oxide.     Calorific 
power   3,062   Calories;   product   one    cubic   meter   of   carbonic 
oxide,  and  1.904  cubic  meters  of  nitrogen. 

Let  t  be  the  temperature  attained ;  then 

Heat  in  the         1  m3  carbonic  oxide  ==  0.37t   +  0.00022t2 

1.904m3  nitrogen  gas       =  0.577t  +  O.Q000514t2 
11       products  =  3,062  Cal's.  =  0.947t  +  0.0002714t2 
Whence  t  =  2050° 

(b)  If  the  2.404  cubic  meters  of  air  needed  are  preheated 
to  700°  C.,  they  will  bring  in  as  sensible  heat 

Q  (o-700)  =  2.404  [0.303  (700) +0.000027  (700)2 
=      552  Calories. 

The    total    heat    in   the    products    will    be    3062  +  552  =  3614 
Calories,  and  therefore  0.947t  + 0.000271 4t2  =  3614 
Whence  t  =  2189° 

(c)  If  the  gas  itself  is  also  preheated  it  brings  in 

Q  (o-700)  =  0.303  (700) +0.000027  (700)2 
=      225  Calories. 

The   total    heat    in   the    products   will    be    3614  +  225  =  3839 
Calories,  and  therefore  0.947t  +  0.0002714t2  =  3839 
Whence  t  =  2284° 

Heating  both  gas  and  air  to  700°  before  they  burn  thus 
raises  the  theoretical  temperature  from  2050  to  2284  or  234°. 

Problem  5. 

Statement. — The  natural  gas  of  Kokomo,  Ind.,  contains  by 
analysis:  Methane  (marsh  gas)  94.16,  ethylene  (olefiant  gas) 


54 


METALLURGICAL  CALCULATIONS. 


0.30,  hydrogen  1.42,  carbonous  oxide  0.55,  carbonic  oxide  0.27, 
oxygen  0.32,  nitrogen  2.80,  hydrogen  sulphide  0.18, — in  per- 
centages, by  volume. 

Required: 

(1)  The  maximum  flame  temperature,  if  burned  cold  with 
the  theoretical  amount  of  cold,  dry  air  necessary. 

(2)  The  calorific  intensity,  if  burned  cold,  with  the  requisite 
air  preheated  to  1000°  C. 

(3)  The  calorific  intensity,  if  burned  cold,  with  25  per  cent, 
more  air  than  theoretically  necessary,  preheated  to  1000°. 

Solution : 

[The  practical  calorific  power  of  this  gas  has  been  already 
calculated  in  Problem  4  as  8203  Calories  per  cubic  meter. 
The  gas  itself  is  always  burned  cold,  because,  if  preheated,  it 
decomposes  and  deposits  carbon  in  the  regenerators.] 

The  products  of  combustion  of  the  gas,  say  per  cubic  meter, 
must  first  be  found,  using  the  formula?  for  combustion. 


Requirement  (1): 


1  Cubic  Meter 
of  Gas. 

Oxygen 
Needed. 

m3 

CO2 

Products. 
H2O 

CH4 

0 

.9416 

1 

.8832 

0 

.9416 

1 

.8832 

C2H4 

0 

.0030 

0 

.0090 

0 

.0060 

0 

.0060 

H2 

0 

.0142 

0 

.0071 

0.0142 

CO 

0 

.0055 

0 

.00275 

0 

.0055 

.... 

CO2 

0 

.0027 

0 

.0027 

.... 

O2 

0.0032 

0.0032         

•N2 

0 

.0280 

.... 

.... 

.... 

H2S 

0 

.0018 

0 

.0027 

0 

.0018 

SO2 


N2 


0.0280 


0.0018 


1.90155  0.9558     1.9052     0.0018     0.0280 
Air  required    -        9.14  (carrying  in  N2)  =  7-238_ 

Total  N2  77266" 

The  heat  generated  will  exist  as  sensible  heat  in  the  CO2, 
H2O,  SO2  and  N2  of  the  products.  The  mean  specific  heat  of 
SO2  per  cubic  meter  is  0.444  at  ordinary  temperatures;  what 
it  is  at  high  temperature  has  not  been  determined;  we  will 
give  it  the  same  index  of  increase  as  the  analogous  gas  CO2, 


THERMOCHEMICA  L  DATA.  55 

and   take   for   it    Sm    (o-t)  =  0.444  +  0.00027t,    or   Q    (o-t)  = 
0.444t  +  0.00027t2. 

At  the  temperature  attained  by  combustion,  t,  the  products 
will  contain  the  following  amounts  of  heat: 

N2  =  7.266  (0.303t  +  0.000027t2) 

H2O  =  1.9052  (0.34t   +0.000l5t2) 

CO2  =  0.9558  (0.37t   +  0.00027t2)* 

SO2  =  0.0018  (0.444t  +  0.00027t2) 

Total  =  3.2044t  +  0.00074057t2  =  8203  Calories. 
From  which  t  =  1806°     (1) 

Requirement  (2): 

Heat  in      1  m3  of  air  at  1000°  =  0.303  (1000)  +0.000027  (1000)2. 

=  330  Calories. 

"      9.14m3      "          1000°  =  3016       " 
Therefore :  3.2044t  +  0.00074057t2  =  8203  +  3016 
Whence  t  =  2288° 

Requirement  (3): 

Excess  air  used  =  9.14x0.25.=  2.285m3. 

Heat  in  11.425m3  at  1000°  =  11.425x330  =  3770  Calories, 

The  heat  capacity  of  the  excess  air  must  be  added  to  the  heat 
in  the  products  at  temperature  t,  viz.: 

Excess  air  =  2.285  (0.303t  +  0.000027t2) 

making  the  total  heat  in  the  products  (adding  to  previous  ex- 
pression) : 

3.8968t  +  0.00080227t2  =  8203  +  3770 
Whence  t  =  2134° 

THE  ELDRED  PROCESS  OF  COMBUSTION. 

A  means  of  regulating  the  temperature  of  the  flame  has 
been  proposed  by  Eldred,  and  described  by  Mr.  Carlton  Ellis 
in  the  December,  1904,  issue  of  El.  Chem.  Ind.  The  proposition 
is  simply  to  mix  with  the  air  used  for  combustion  a  certain 

*  These  and  some  succeeding  problems  were  worked  out  using  this 
specific  heat.  Later  the  author  has  adopted  a  slightly  different  value  for 
CO2,  viz.,  (0.37  +  0.00022t),  which  he  now  considers  more  nearly  correct. 


56  METALLURGICAL  CALCULATIONS. 

proportion  of  the  products  of  combustion  themselves.  The 
principle  is  easily  understood  when  the  requisite  calculations 
of  the  theoretical  flame  temperatures  are  made.  When  solid 
fuel  is  burnt  the  temperature  is  often  too  high  locally,  and  re- 
sults in  burning  out  grate  bars  or  overheating  the  brick  work 
of  the  fire-place,  or  overheating  locally  the  material  which  is 
mixed  with  the  fuel.  If  the  air  is  diluted  with  products  of 
combustion  the  initial  theoretical  temperature  is  lowered,  and 
the  above  evils  may  be  obviated.  Using  solid  fuel,  the  heat  in 
the  fuel  before  the  air  actually  burns  it  must  be  added  to  the 
heat  generated  by  combustion  to  get  the  actual  temperature  in 
the  hottest  part  of  the  fire. 

Examples:  (1)  What  will  be  the  highest  temperature  in  a 
charcoal  fire  fed  by  air,  assuming  complete  combustion  without 
excess  of  air? 

Assuming  the  charcoal  to  be  pure  carbon,  and  to  be  heated 
to  the  maximum  temperature  t  before  it  burns  (by  the  com- 
bustion of  the  preceding  part),  the  heat  available  is: 

Heat  of  combustion  of  1  kilo  of  carbon 8100  Cal's. 

Sensible  heat  in  the  carbon  at  t°. .  .  .0.5t— 120      " 


Total  heat  available  to  raise  the  temperature. .  .7980  +  0.5t 

Products  of  combustion CO2  1.85  m3 

..N2     7.04  " 

Heat  of  product  at  t° CO2  1.85  (0.37t  +0.00022t2) 

N2     7.04  (Q.303t  +  O.OQ0027t2) 

Sum  2.81t  +  0.00060t2 
therefore     2.81t  +  0.00060t2  =  7980+0.5t 
whence  t  =  2199° 

(2)  What  will  be  the  temperature  in  the  same  case,  if  the 
air  used  is  diluted  with  an  equal  volume  of  the  products  of 
combustion  ? 

The  heat  available  is  the  same  as  before,  7980  +  0.5t;  but 
since  the  mixed  air  for  combustion  contains  only  half  as  much 
oxygen  per  cubic  meter,  the  products  will  be  exactly  doubled  in 
amount  for  a  unit  weight  of  carbon  burnt,  and  we  therefore 
have  directly 

2  (2.810t  +  0.00060t2)  =  79SO  +  0.5t 
whence  t  =  1213° 


THERMOCHEMICA  L  DA  TA  .  57 

It  is  therefore  evident  that  the  maximum  temperature  of  the 
hot  gases  at  their  moment  of  formation  is  nearly  halved  by  the 
procedure  stated. 

(3)  Taking  the  cases  cited  by  Mr.  Ellis,  where  the  products 
contained  originally  15  per  cent,  oxygen  and  6  per  cent,  carbon 
dioxide,  and  after  mixing  the  air  used  with  half  its  volume  of 
the  chimney  gases,  9  per  cent,  oxygen  and  12  per  cent,  carbon 
dioxide  (the  gas-air  mixture  entering  containing  15  per  cent. 
oxygen  and  6  per  cent,  carbon  dioxide),  what  are  the  maximum 
temperattires  obtained  in  the  two  cases? 

Case  1  :  The  heat  available  is  as  before  ;  the  products  of  com- 
bustion are  CO2  1.85  m3,  O2  4.62  m3,  N2  24.36  m3,  and  their 
sensible  heat  at  temperature  t  — 

Heat  in  CO2        =     1.85  (0.37t   +0.00027t2) 
=  28.98  (0.303t  +  0.000027t2) 


Sum       «         9.46t  +  0.00128t2 

therefore  9.46t  +  0.001280t2  =  7980  +  0.5t 
whence  t  =  800° 

Case  2:  The  products,  per  kilogram  of  carbon  burnt,  will 
be  CO2  3.70  m3,  O2  2.77  m3,  N2  24.36  m3,  and  we  have 

Heat  in  CO2        =    3.70  (0.37t   +0.00027t2) 
O2  +  N2  =  27.13  (0.303t  +  0.000027t2) 

Sum     =  "      9.09t  +  0.00173t2 

therefore  9.09t  +  0.00173t2  =  7980  +  0.5t 
whence  t  =  764° 

Conclusions:  The  calculations  regarding  the  Eldred  process 
bring  out  what  was  not  stated  in  the  printed  description  of 
the  method,  viz.:  that  if  a  small  excess,  or  no  excess  of  oxygen 
escapes,  to  the  chimney,  the  temperature  of  the  flame  will  be 
greatly  reduced  by  the  dilution  of  the  air  used,  because  more 
gas-air  mixture  will  be  required  per  unit  of  fuel  burnt;  but  if 
there  is  any  large  amount  of  unused  oxygen  escaping  in  the 
first  instance,  the  dilution  practiced  will  scarcely  affect  the 
temperature  of  the  flame  at  all,  because  it  makes  very  little 
difference  whether  the  fuel  is  heating  up  unused  oxygen  or 
carbon  dioxide  dilutant  substituted  for  some  of  it,  as  long 


58  METALLURGICAL  CALCULATIONS. 

is  there  is  more  than  enough  oxygen  to  burn  the  fuel.  The 
fact  of  the  specific  heat  of  carbon  dioxide  being  greater  than 
that  of  oxygen,  is  the  only  reason  for  the  small  calculated  dif- 
ference of  36°,  in  cases  1  and  2.  It  is  true  that  the  dilution 
practiced  will  result  in  less  heat  being  lost  in  the  waste  gases, 
in  the  Example  (3),  but  the  same  economy  could  be  obtained 
by  simply  using  less  excess  of  air  in  the  first  instance. 

TEMPERATURES  IN  THE  "  THERMIT  "  PROCESS. 

The  Goldschmidt  porcess  of  reducing  metallic  oxides  by 
powdered  aluminium,  igniting  the  cold  mixture,  is  only  a  spe- 
cial case  of  our  general  rule,  as  far  as  concerns  the  calcula- 
tion of  the  theoretical  temperatures  obtained.  In  any  case, 
the  total  heat  available  is  the  surplus  evolved  in  the  chemical 
reaction,  and  the  temperature  sought  is  that  to  which  this 
quantity  of  heat  will  raise  the  products  of  the  reduction.  The 
products  are  alumina  and  the  reduced  metal.  The  heat  in  the 
latter,  in  the  melted  state,  is  well  known  in  many  cases;  it 
is  most  clearly  expressed  by  the  sum  of  the  heat  in  such  melted 
metal  just  at  its  melting  point  (which  is  easily  determined 
calorimetrically,  and  is  well  known  for  many  metals),  plus 
the  heat  in  the  melted  metal  from  the  melting  point  to  its 
final  temperature,  which  is  equal  to  t,  minus  the  melting  tem- 
perature, multiplied  by  the  specific  heat  in  the  melted  condition 
These  data  are  known  for  many  metals;  for  some  they  may 
have  to  be  assumed  from  some  general  laws  correlating  these 
values.  The  heat  in  melted  alumina  has  not  been  determined 
calorimetrically,  to  my  knowledge.  Its  sensible  heat  solid  is 
0.2081t  +  0.0000876t2  (determination  made  in  author's  labora- 
tory), which,  evaluated  for  the  probable  melting  point,  2200° 
C.,  would  give  the  heat  in  it  at  that  temperature  881.8  Calories; 
the  latent  heat  of  fusion  of  molecular  weight  is  very  probably 
2.1  T,  where  T  is  the  absolute  temperature  of  the  melting  point, 
making  the  latent  heat  of  fusion  per  kilogram  2.1  (2200  + 
273) -^  102  =  5193-^102  =  50.9  Calories.  The  specific  heat  in 
the  melted  state  is  probably  equal  to  the  specific  heat  of  the 
solid  at  the  melting  point,  viz.:  0.2081+0.0001752  (2200)  = 
0.5935.  We,  therefore,  have  for  the  heat  in  melted  alumina  at 
temperature  t — 


THERMOCHEMICAL  DATA.  59 

Heat  in  solid  alumina  to  the  melting  point 881 .8  Calories 

Latent  heat  of  fusion 50.9 

Heat  in  liquid  alumina  to  its  setting  point 0.5935  (t-2200) 

Total. 932. 7 +  0.5935  (t-2200) 
or,  fora  molecular  weight,  102  kilograms:  95,135  +  60.54  (t-2200). 

Examples: 

(1)  If  black  cupric  oxide  is  reduced  by  powdered  aluminium 
what  is  the  temperature  attained? 

The  react  ion  "is 

3CuO  +  2Al  =  Al2O3  +  3Cu 
—3  (37,700)        '  +  392,600 


+  279,500 

The  products  must  therefore  be  raised  to  such  a  tempera- 
ture that  they  contain  279,500  Calories.  The  heat  in  molecular 
weight  of  alumina  at  temperature  t  has  already  been  found; 
that  in  copper  at  t°  is,  for  one  kilogram  (using  the  author's 
determinations) : 

Heat  in  melted  copper  at  setting  point  =162  Calories. 

Heat  in  melted  copper  above  1065°  =  0.1318  (t— 1065) 
Calories. 

Total  =  162  +  0.1318  (t— 1065). 

Per  atomic  weight  (63.6  kilos.)  =  10303  +  8.3825  (1—1065) 

From  these  data  there  follows  the  equation: 

95,135  +  60.54  (t— 2200)  +  3[10,303  + 

8.3825  (t— 1065)]  =  279,500 
whence  t  =  3670° 

A  little  further  calculating  will  show  that  approximately  one- 
third  of  all  the  heat  generated  exists  in  the  copper,  and  two- 
thirds  in  the  melted  alumina. 

(2)  If  pure  ferric  oxide  is  reduced  by  the  "  Thermit  "  process, 
what  is  the  temperature  of  the  resulting  iron  and  melted  alumina  ? 

The  reaction  and  the  heat  evolved  have  been  already  given 
in  the  preceding  discussion  of  these  calculations  (page  44.) 
They  show  that  per  molecular  weight  of  alumina  formed  there 
are  two  atomic  weights  (112  kilos.)  of  iron  formed,  and  there 
is  disposable  altogether  197,000  Calories.  The  heat  in  a  kilo- 


60  METALLURGICAL  CALCULATIONS. 

gram  of  pure  iron  at  its  melting  point  (1600°)  is  300  Calories, 
the  latent  heat  of  fusion  approximately  69  Calories,  and  the 
specific  heat  in  the  melted  condition  0.25.  The  total  heat  in  a 
kilogram  of  melted  iron  is  therefore  369  +  0.25  (t— 1600)  Cal- 
ories, or  per  atomic  weight  =  20,664  +  14  (t— 1600)  Calories. 
We,  therefore,  can  write  the  equation: 

95,135  +  60.54  (t— 2200) +  2[20,664  +  14  (t— 1600)]  =  197,000 
whence  t  =  2694° 

A  similar  calculation  made  for  the  reduction  of  MnO  by  the 
theoretical  amount  of  aluminium,  shows  a  reduction  tem- 
perature less  than  the  melting  point  of  alumina.  This  would 
mean  that  the  melting  down  of  the  mass  to  a  fused  slag  of 
pure  alumina  could  not  take  place.  What  happens  in  the  re- 
duction of  manganese  is  that  an  excess  of  manganous  oxide  is 
used,  whereby  all  the  aluminium  is  consumed,  and  none  at  all 
gets  into  the  reduced  manganese,  and,  furthermore,  the  excess 
of  manganous  oxide  unites  with  the  alumina  to  form  a  slag 
of  manganous  aluminate,  which  is  fusible  at  the  temperature 
attained.  Without  the  latter  arrangement  no  fused  slag  could 
result. 

Similar  calculations,  made  with  silicon  as  the  reducing 
agent,  show  similar  difficulties  regarding  the  theoretical  tem- 
peratures attainable,  when  iron  or  manganese  are  reduced. 
By  using  an  excess  of  the  oxides  of  these  metals,  however, 
calculation  shows  that  temperatures  sufficient  to  fuse  the 
metals  and  the  manganous  silicate,  or  ferrous  silicate  produced, 
ought  to  be  obtained. 


CHAPTER  TV. 

THE  THERMOCHEMISTRY  OF  HIGH  TEMPERATURES 

The  problem  is:  Knowing  the  heat  evolved  (or  absorbed) 
in  the  formation  of  a  compound,  or  in  a  double  reaction,  start- 
ing with  the  reacting  materials  cold,  and  ending  with  the  pro- 
ducts cold,  what  is  the  heat  evolved  (or  absorbed)  in  either  of 
the  following  cases? 

(a)  Starting  with  the   reagents   cold  and  ending  with  the 
products  hot. 

(b)  Starting   with  the    reagents   hot   and   ending   with  the 
products  hot. 

(c)  Starting  with  the  reagents  hot  and  ending  with  the  pro- 
ducts cold. 

Of  these  three  cases  (b)  is  the  most  general  form  of  the  prob- 
lem, and  occurs  frequently  in  metallurgical  practice,  particu- 
larly in  electrometallurgy;  (a)  is  a  more  limited  form,  and 
requires  less  data  for  its  calculation,  while  it  is  very  frequently 
the  desideratum  in  discussing  the  thermochemistry  or  heat  re- 
quirements of  a  metallurgical  process;  (c)  is  derivable  at  once 
if  the  data  exist  for  calculating  (b),  and  is  of  such  rare  occur- 
rence in  practice  that  we  can  dispense  with  its  lengthy  dis- 
cussion. 

The  thermochemical  data  already  given  and  described  in  pre- 
ceding sections  enable  us  to  calculate  the  heat  of  any  chemical 
reaction  starting  with  cold  reagents  and  ending  with  the  pro- 
ducts cold.  For  instance:  (Zn,  O)  =  84,800  Calories  means 
that  if  we  take  65  kilograms  of  solid  zinc,  at  room  tempera- 
ture, and  16  kilograms  of  oxygen,  as  gas  at  room  temperature, 
ignite  them,  and  after  the  reaction  cool  the  81  kilograms  of 
zinc  oxide  formed  down  to  the  same  starting  temperature, 
there  will  be  developed  a  total  of  84,800  Calories.  Similarly, 
using  the  datum  (C,  0)  =  29,160  Calories,  we  can  construct 

61 


60  METALLURGICAL  CALCULATIONS. 

gram  of  pure  iron  at  its  melting  point  (1600°)  is  300  Calories, 
the  latent  heat  of  fusion  approximately  69  Calories,  and  the 
specific  heat  in  the  melted  condition  0.25.  The  total  heat  in  a 
kilogram  of  melted  iron  is  therefore  369  +  0.25  (t — 1600)  Cal- 
ories, or  per  atomic  weight  =  20,664+14  (t— 1600)  Calories. 
We,  therefore,  can  write  the  equation: 

95,135  +  60.54  (t— 2200) +  2[20,664+14  (t— 1600)]  =  197,000 
whence  t  =  2694° 

A  similar  calculation  made  for  the  reduction  of  MnO  by  the 
theoretical  amount  of  aluminium,  shows  a  reduction  tem- 
perature less  than  the  melting  point  of  alumina.  This  would 
mean  that  the  melting  down  of  the  mass  to  a  fused  slag  of 
pure  alumina  could  not  take  place.  What  happens  in  the  re- 
duction of  manganese  is  that  an  excess  of  manganous  oxide  is 
used,  whereby  all  the  aluminium  is  consumed,  and  none  at  all 
gets  into  the  reduced  manganese,  and,  furthermore,  the  excess 
of  manganous  oxide  unites  with  the  alumina  to  form  a  slag 
of  manganous  aluminate,  which  is  fusible  at  the  temperature 
attained.  Without  the  latter  arrangement  no  fused  slag  could 
result. 

Similar  calculations,  made  with  silicon  as  the  reducing 
agent,  show  similar  difficulties  regarding  the  theoretical  tem- 
peratures attainable,  when  iron  or  manganese  are  reduced. 
By  using  an  excess  of  the  oxides  of  these  metals,  however, 
calculation  shows  that  temperatures  sufficient  to  fuse  the 
metals  and  the  manganous  silicate,  or  ferrous  silicate  produced, 
ought  to  be  obtained. 


CHAPTER  TV. 

THE  THERMOCHEMISTRY  OF  HIGH  TEMPERATURES. 

The  problem  is:  Knowing  the  heat  evolved  (or  absorbed) 
in  the  formation  of  a  compound,  or  in  a  double  reaction,  start- 
ing with  the  reacting  materials  cold,  and  ending  with  the  pro- 
ducts cold,  what  is  the  heat  evolved  (or  absorbed)  in  either  of 
the  following  cases? 

(a)  Starting  with  the   reagents   cold   and  ending  with  the 
products  hot. 

(b)  Starting   with   the    reagents   hot   and   ending   with   the 
products  hot. 

(c)  Starting  with  the  reagents  hot  and  ending  with  the  pro- 
ducts cold. 

Of  these  three  cases  (b)  is  the  most  general  form  of  the  prob- 
lem, and  occurs  frequently  in  metallurgical  practice,  particu- 
larly in  electrometallurgy;  (a)  is  a  more  limited  form,  and 
requires  less  data  for  its  calculation,  while  it  is  very  frequently 
the  desideratum  in  discussing  the  thermochemistry  or  heat  re- 
quirements of  a  metallurgical  process;  (c)  is  derivable  at  once 
if  the  data  exist  for  calculating  (6),  and  is  of  such  rare  occur- 
rence in  practice  that  we  can  dispense  with  its  lengthy  dis- 
cussion. 

The  thermochemical  data  already  given  and  described  in  pre- 
ceding sections  enable  us  to  calculate  the  heat  of  any  chemical 
reaction  starting  with  cold  reagents  and  ending  with  the  pro- 
ducts cold.  For  instance:  (Zn,  O)  =  84,800  Calories  means 
that  if  we  take  65  kilograms  of  solid  zinc,  at  room  tempera- 
ture, and  16  kilograms  of  oxygen,  as  gas  at  room  temperature, 
ignite  them,  and  after  the  reaction  cool  the  81  kilograms  of 
zinc  oxide  formed  down  to  the  same  starting  temperature, 
there  will  be  developed  a  total  of  84,800  Calories.  Similarly, 
using  the  datum  (C,  O)  =  29,160  Calories,  we  can  construct 

61 


62  METALLURGICAL  CALCULATIONS. 

and  interpret  the  reduction  reaction,   starting  with  cold  ma- 
terials and  finally  ending  with  cold  materials,  as  follows: 

ZnO  +  C  =  Zn  +  CO 
—84,800          I          +29,160 


—55,640 

This  reaction,  as  interpreted,  stands  for  none  of  the  above 
(a),  (b)  or  (c) ;  in  fact,  it  represents  only  a  calorimetric 
determination  in  the  laboratory,  and  does  not  correspond  to 
either  of  the  three  cases  actually  taking  place  in  practice,  that 
is,  it  is  not  directly  applicable  to  practical  conditions,  without 
being  modified  by  the  conditions  actually  arising  in  practice. 

Case  (a):  If  we,  in  practice,  start  with  the  reagents  cold, 
and  the  products  pass  away  from  the  furnace  hot,  at  some  de- 
termined temperature  t,  the  total  heat  energy  necessary  to 
cause  this  transformation  is  calculable  in  two  ways.  The  first 
way  is  to  follow  the  course  of  the  reaction,  and  to  say  that 
the  total  heat  absorbed  is  that  necessary  to  heat  the  reacting 
bodies  to  the  temperature  t,  plus  the  heat  of  the  chemical  re- 
action assumed  as  starting  and  finishing  at  that  temperature. 
The  first  of  these  items  can  be  obtained  if  we  know  the  sensi- 
ble heat  in  the  reacting  bodies  up  to  the  temperature  t ;  it  is 
a  question  of  specific  heats  of  the  reacting  bodies  up  to  t  (in- 
cluding any  physical  changes  of  state  occurring  in  them  be- 
tween o  and  t);  the  second  item  requires  a  knowledge  of  the 
heat  of  formation  of  all  the  compounds  involved,  starting  with 
their  ingredients  at  t,  and  ending  with  the  products  at  t.  But 
the  latter  item  is  the  general  question  of  the  heat  of  a  reaction 
starting  with  the  ingredients  hot  and  ending  with  the  pro- 
ducts hot;  it  is  the  most  general  case,  which  we  have  desig- 
nated as  Case  (6),  and  which  will  be  discussed  later.  This 
way  of  solving  Case  (a),  therefore,  really  includes  the  solu- 
tion of  Case  (b),  and  we  will  defer  its  consideration  for  the 
present.  The  second  method  of  solving  Case  (a),  and  one 
which  does  not  involve  the  more  general  solution,  is  to  take 
the  heat  of  the  reaction,  starting  with  the  ingredients  cold  and 
ending  with  the  products  cold — the  ordinary  heat  of  the  re- 
action from  ordinary  thermochemical  data — and  to  add  to  this 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         63 

the  amount  of  heat  which  would  be  required  to  raise  the  pro- 
ducts from  zero  to  the  temperature  t.  This,  of  course,  does 
not  actually  represent  the  sequence  in  which  the  reactions  take 
place  in  practice,  but  it  accurately  represents  the  heat  involved 
or  evolved  in  passing  from  the  cold  reagents  to  the  hot  pro- 
ducts, and  is,  therefore,  exactly  the  practical  quantity  which 
we  are  endeavoring  to  find.  Moreover,  it  involves  a  knowledge 
of  only  the  specific  heats  of  the  products,  and  not  that  of  the 
ingredients  or  substances  reacting. 

Illustration:  Starting  with  a  mixture  of  zinc  oxide  and  car- 
bon in  the  proportions  Zn  O  and  C,  at  ordinary  temperature, 
and  shoveling  them  cold  into  a  retort,  how  much  heat  is  ab- 
sorbed in  converting  them  into  Zn  vapor  and  CO  gas,  issuing 
from  the  retort  at  1300°  C.? 

If  we  start  to  calculate  this  quantity,  by  first  finding  the  heat 
necessary  to  heat  Zn  O  and  C  up  to  1300°,  that  is  obtained  by 
multiplying  the  weights  of  each  by  their  mean  specific  heats 
from  0°  to  1300°,  and  then  by  1300,  as  follows: 

Calories. 

81  kilos.  ZnOX*[0.1212  (1300) +0.0000315  (1300)2]  =  17,075 
12     "      C      X*[0.5  (1300)— 120]  =    6,360 

Sum  =  23,435 

To  this  must  then  be  added  the  heat  of  the  reaction  ZnO  + 
C  =  Zn  +  CO,  starting  with  the  reacting  bodies  at  1300°,  and 
ending  with  the  products  at  the  same  temperature.  This  can 
only  be  found  by  solving  the  general  Case  (b)  for  this  par- 
ticular reaction,  which  we  will  find  involves  a  knowledge  of  the 
heat  required  to  raise  Zn,  O,  C.  ZnO  and  CO  from  0°  to  t. 
Anticipating  such  a  solution,  we  may  say  that  the  heat  of  the 
reaction  starting  and  ending  at  1300°  is — 80,166  Calories, 
making  the  sum  total  of  energy  required  103,601  Calories. 

The  solution  is  usually  much  simpler  if  we  take  the  second 
method,  and  add  to  the  heat  of  the  reaction,  starting  and 
ending  at  zero  ( — 55,640  Calories),  the  heat  required  to  raise 
65  kilos,  of  zinc  and  28  kilos,  (22.22  cubic  meters)  of  carbonous 


*Heat  in  1  kilo  of  carbon,  for  temperature  above  1,000°,  0.5t  -  120 
(deduction  from  Weber's  results) ;  zinc  oxide  0.1212  t  +  0-0000315 12  (de- 
termination by  the  author). 


64  METALLURGICAL  CALCULATIONS. 

oxide,  from    their    ordinary  condition  at  zero  to  their  normal 
condition  at  1300°.     The  calculation  for  the  CO  gas  is  simply: 

22.22X[0.303  (1300)4-0.000027  (1300)2]  =  9,766  Calories. 

For  the  zinc,  the  calculation  is  more  complicated: 
Heat  in  solid  zinc  to  the  melting  point 
(420°) : 

65X[0.09058  (420) +0.000044  (420)2]  =  2,977  Calories. 
Latent  heat  of  fusion  65X22.61   =1,470         " 

Heat  in  melted   zinc,   420°  to  boiling  point 
(930°) : 

65 X [0.0958 +  0.000088    (420)]X   (930—420)      =4,228        " 
Latent  heat  of  vaporization  (Trouton's  rule) 

23 X  (930  +  273)   =  27,670       u 
Heat  in  zinc  vapor   (monatomic)   5X    (1300 

—930)   =  1,850  Calories. 

Sum  =  38,195       " 

The  total  sensible  heat  required  is,  therefore,  9,766  +  38,195 
=  47,961  Calories,  which,  added  to  the  55,640  absorbed  in  the 
chemical  reaction,  if  it  started  and  ended  at  zero,  makes  a  total 
heat  requirement  of  103,601  Calories  for  the  practical  carrying 
out  of  this  reaction,  starting  with  the  reagents  cold  and  ending 
with  the  hot  products  at  1300°. 

[To  be  absolutely  accurate,  regard  should  be  paid  in  the 
above  case  to  the  fact  that  the  above  calculations  are  based 
on  the  substances  being  all  at  atmospheric  pressure,  while  in 
the  mixture  of  Zn  vapor  and  CO  gas  each  is  under  only  0.5 
atmospheric  tension.  Since  each  of  these  represents  a  molecular 
weight,  the  outer  work  which  has  been  included  in  the  calcu- 
lations is  2X2T  =  2X2  (1300  +  273)  =  6292  Calories,  whereas 
it  should  really  be  only  half  that  much,  or  3146  Calories.  The 
corrected  heat  required  is,  therefore,  103,601—3,146  =  100,455 
Calories,  or  1,545  Calories  per  kilogram  of  zinc.  This  datum 
is  exactly  the  net  heat  requirement  on  which  calculations  of 
the  net  electrical  energy  required  to  produce  zinc  from  its  oxide, 
or  calculations  of  the  net  efficiency  of  an  ordinary  zinc  furnace, 
would  be  based.] 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        65 

Case  (6):  To  calculate  the  heat  of  a  chemical  reaction  start- 
ing and  finishing  at  any  temperature  t,  two  methods  are  avail- 
able; The  most  general  solution,  and  that  easiest  to  understand, 
is  to  calculate  for  each  compound  involved  the  heat  of  its 
formation  at  the  temperature  t,  that  is,  the  heat  evolved  if 
the  elements  start  at  t  and  the  product  is  cooled  to  t.  Having 
these  heats  of  formation  at  t,  they  are  used  in  the  equation 
in  just  the  same  manner  as  the  heats  of  formation  at  zero 
are  ordinarily  used  in  obtaining  the  heat  of  the  reaction  start- 
ing and  ending  at  zero.  The  calculations  are  based  on  this 
general  principle:  The  heat  evolved  when  the  cold  elements 
unite  to  form  the  hot  product  at  temperature  t  equals  the  heat 
of  union  at  zero,  minus  the  heat  necessary  to  raise  the  product 
from  zero  to  t ;  if  to  this  difference  we  add  the  heat  which  would 
be  necessary  to  heat  the  uncombined  elements  from  zero  to  t, 
the  sum  is  the  desired  heat  of  formation  at  t°. 

Illustration:  The  heat  of  formation  of  ZnO  at  zero  is  84,800, 
starting  with  cold  Zn  and  O  and  ending  with  cold  ZnO.  If 
we  started  with  cold  Zn  and  O  and  ended  with  hot  ZnO,  say 
at  1300°,  the  heat  evolved  altogether  would  be  less  than  84,800 
by  the  sensible  heat  in  the  81  kilograms  of  ZnO  at  1300°,  which 
has  already  been  calculated  (see  previous  illustration)  to  be 
17,075  Calories.  The  difference  is  67,725  Calories,  and  repre- 
sents the  transformation  from  cold  Zn  and  O  to  hot  ZnO. 
If  the  Zn  and  O  were  heated  to  1300°  before  combining,  they 
would  contain  as  sensible  heat  the  following  quantities: 

Heat  in  65  kilograms  of  Zn  (0  to  1300°) 

already  calculated  38,195  Calories 
Heat  in  16  kilograms  of  O  =  11.11  m3X 

[0.303  (1300) +0.000027  (1300)2]  =     4,884 

Sum     43,079 

And  the  heat  evolved  in  passing  from  the  hot  reagents  to  the 
hot  product  at  1300°  must  be  67,725  plus  this  sensible  heat, 
or  110,804  Calories.  We  can  express  this  datum  as  follows: 
(Zn,  O)1300  =  110,804,  which  means  that  when  the  zinc  and 
oxygen  taken  in  their  normal  state  at  1300°  combine  to  form 
ZnO  at  1300°,  110,804  Calories  are  evolved. 


66  METALLURGICAL  CALCULATIONS. 

A  similar  calculation  for  CO  is  as  follows: 

(C,  O)  =  29, 160  Calories 

Sensible  heat  in  CO  (0  to  1300°)  already  cal- 
culated =     9,766 

Heat  evolved  when  cold  C  and  O  form  CO 

at  1300°  =  19,394 
Sensible  heat  in  C  (0  to  1300°)  =  6360 

"  O  (         "         )  =  4884  11,244  Calories 


Heat  evolved  hot  C  and  O  to  hot  CO  =  Sum  =  30,638 
Or  (C,  O)1300  =  30,638 

Uniting  the  two  data  found,  for  the  heats  of  formation  of 
ZnO  and  CO  at  1300°,  in  the  equation  of  reduction,  we  have 

ZnO  +  C  =  Zn  +  CO 

at  1300  °  =  -110,804    |          +30,638 

—80,166 

The  actual  reduction  is  thus  seen  to  absorb  24,526  Calories 
more  at  1300°  than  at  zero,  an  increase  of  over  42  per  cent. 

If  to  this  heat  of  reaction  at  1300°  we  add  the  heat  neces- 
sary to  raise  the  ZnO  and  C  to  1300°  (already  calculated  under 
Case  (a)  as  23,435  Calories),  we  will  have  a  total  heat  require- 
ment of  103,601  Calories,  which  would  be  required  practically 
if  we  started  with  cold  ZnO  and  C  and  ended  with  the  hot 
Zn  and  CO.  This  agrees,  as  it  indeed  must,  with  the  sum 
of  the  heat  absorbed  in  the  reaction  at  zero,  55,640,  increased 
by  the  sensible  heat  in  Zn  and  CO  at  1300°,  already  found  to 
be  47,961,  or  a  total  of  103,601. 

Another  method  of  calculating  the  heat  of  the  reaction 
ZnO  +  C  =  Zn  +  CO  at  1300°,  without  using  the  heat  of  forma- 
tion of  ZnO  and  CO  at  1300°,  is  based  on  the  following  gen- 
eral principle:  If  from  the  heat  of  any  reaction,  starting  and 
ending  cold,  there  be  subtracted  the  heat  necessary  to  raise 
the  products  from  0  to  t,  the  difference  is  the  heat  of  the  trans- 
formation from  cold  reagents  to  hot  products;  if  to  this  be 
added  the  heat  which  would  be  contained  in  the  reagents  if 
they  were  heated  to  t,  the  sum  is  the  heat  of  transformation 
from  heated  reagents  to  heated  products,  all  at  the  tempera- 
ture. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         67 

Illustration:  The  heat  of  the  reaction  we  have  been  study- 
ing, starting  and  ending  cold,  is  — 55,640  Calories 
Sensible  heat  in  Zn  at  1300°  =  38,195 

"     CO        "  =    9,766     47,961 


Difference— 103,601 
Sensible  heat  in  ZnO  at  1300°  =  17,075 

C  =    6,360     23,435        " 

Sum— 80,166 

The  above  is  the  simplest  way  of  calculating  the  heat  of  any 
chemical  reaction  at  any  desired  temperature,  since  it  involves 
the  knowledge  of  the  heat  capacities  of  only  those  substances 
which  occur  individually  in  the  reaction,  and  not  that  of  the 
elements  of  which  the  compounds  present  are  composed.  The 
above  calculation,  for  instance,  involves  the  heat  capacities 
of  ZnO,  C,  Zn,  and  CO,  but  not  that  of  oxygen,  which  does  not 
occur  free  in  the  reaction. 

Case  (c) :  This  hardly  needs  discussion,  because  if  we  have 
the  data  for  calculating  Case  (b),  this  case  can  be  easily  worked. 
If  to  the  heat  of  the  reaction  at  t°  we  add  the  heat  given  out 
by  the  products  in  cooling  from  t  to  0,  the  sum  is  the  total 
heat  evolved  in  passing  from  the  hot  reagents  to  the  cold  pro- 
ducts. This  solution  of  the  problem  presupposes,  however, 
the  solution  of  Case  (b)  and  requires  the  maximum  amount 
of  data,  but  it  has  the  advantage  of  following  and  representing 
the  logical  course  of  the  reaction.  A  simpler  solution  is  to  add  to 
the  heat  of  the  cold  reaction  at  zero,  the  heat  necessary  to  heat 
the  ingredients  to  t,  and  the  sum  will  be  the  quantity  required. 
Illustration :  Taking  the  same  case  as  before : . 

Heat  of  reaction  at  1300°  =  —80,166 

Sensible  heat  in  products  at  1300°          =     ,  47,961 


Sum  =  —32,205 


or,        Sensible  heat  in  reagents  at  1300°  =       23,435 

Heat  of  reaction  at  zero  =  — 55,640 

Sum  =  —32,205 


The  reasoning  involved  in  the  above  is  simply  that,  starting 
with  the  reagents  hot  and  ending  with  the  products -cold,  the 


68  METALLURGICAL  CALCULATIONS. 

heat  evolution  must  be  the  same  whether  we  suppose  the  sys- 
tem to  pass  along  the  one  path  or  the  other. 

GENERAL  REMARKS. 

It  will  be  evident  from  this  enunciation  of  principles  and  the 
illustrations  adduced,  that  for  many  practical  purposes  the 
calculation  according  to  Case  (a)  will  suffice  for  finding  the 
net  energy  involved  in  many  metallurgical  operations;  it  gives 
the  sum  total  of  energy  necessary  to  be  supplied  to  pass  from 
the  cold  reagents  to  the  hot  products  as  they  come  from  the 
furnace,  but  the  two  items  of  which  this  sum  is  composed  do 
not  actually  represent  the  two  items  of  the  division  of  this 
total  in  the  furnace,  i.e.,  into  heat  necessary  to  heat  the  re- 
agents to  the  reacting  temperature  and  heat  actually  absorbed 
as  the  reaction  takes  place.  The  latter  item  can  only  be  calcu- 
lated by  one  of  the  two  methods  explained  under  Case  (b)t 
and  then  the  former  item  can  be  obtained  by  difference,  or  by 
direct  use  of  the  heat  capacities  of  the  reacting  substances. 

It  is,  of  course,  obvious  that  this  whole  subject  of  calcu- 
lating the  thermochemistry  of  high  temperatures  (as  distin- 
guished from  the  ordinary  zero  thermochemistry}  necessitates 
the  use  of  all  available  data  regarding  specific  heats  in  the 
solid,  liquid  and  gaseous  states  of  both  elements  and  com- 
pounds, and  also  in  many  cases  of  their  latent  heats  of  fusion 
and  vaporization.  When,  however,  the  necessary  physical  data 
are  known,  or  can  be  assumed  with  approximate  accuracy,  the 
way  is  open  to  make  many  calculations  of  the  greatest  value  in 
practical  metallurgy  and  chemistry.  The  exact  heats  of  forma- 
tion of  chemical  compounds  at  a  certain  temperature,  which 
may  be,  and  often  are,  very  different  from  these  values  at  zero, 
and,  having  frequently  different  relations  to  each  other,  will 
explain  in  many  cases  many  hitherto  little  understood  and  ap- 
parently contradictory  reactions.  The  calculations  also  enable 
us  to  understand  many  reactions  taking  place  only  at  high  tem- 
peratures, to  calculate  whether  they  are  really  exothermic  or 
endothermic  at  the  temperatures  at  which  they  occur,  and  to 
compare  these  data  with  the  data  as  to  the  heat  of  such  re- 
actions obtained  by  studying  the  chemical  equilibrium  of  such 
reactions  and  deducing  the  heat  of  the  reaction  from  the  rate 
of  displacement  of  the  chemical  equilibrium.  One  such  ex- 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         69 

ample  may  suffice  to  suggest  the  large  field  here  open  to  the 
scientific  metallurgist. 

Example:  G.  Preuner  (Zeitschrift  fur  Physikalische  Chemie* 
March  15,  1904,  p.  385)  reports  a  long  and  careful  investigation 
of  the  equilibrium  of  the  reaction 

Fe3O4  +  4H2  ;=.'  3Fe  +  4H20 

in  which  he  finds  the  equilibrium  constant  for  different  tem- 
peratures, and  calculates  from  its  value  at  960°,  by  the  use  of 
Van't  Kofi's  formula,  that  the  heat  value  of  the  reduction  at 
that  temperature  is — 11,900  Calories,  and  noticing  the  great 
difference  between  this  value  and  the  value  derived  from  the 
ordinary  heats  of  formation  (—270,800  +  4(58,060)  =  —38,560), 
he  concludes  that  the  Van't  Hoff  formula  gives  wrong  re- 
sults when  applied  to  this  class  of  reactions.  Now,  the  facts 
are  that  Preuner's  observations  were  good,  Van't  Kofi's  for- 
mula is  correct,  and  applies,  but  the  thermochemical  value  of 
the  reaction, — 38,560  Calories  is  the  correct  value  only  for 
the  reaction  beginning  and  ending  at  ordinary  temperature.  Our 
thermochemical  principles  enable  us  to  calculate  the  heat  of 
the  reaction  at  960°,  as  follows: 

Heat  of  the  reaction  beginning  and  ending  at 

zero  =  —  38,560  Calories 
Heat  in  products  at  960°: 
3Fe  =  3(56)  X [0.218(960)  —  39]  Pionchon  = 

28,560 
4H20  =  4(22.22)  X [0.34(960) +0.00015. 

(960)2]  =  41,300 

69,860  Calories 
Heat  in  reagents  at  960°: 

Fe3O4  =  232X[0.1447  (960) +0.0001878 
(960)2]*  =  72,384 

4H2  =  4(22.22)  X[0.303(960) +0.000027 
(960)2]  =  28,075 

100,459       " 

The  heat  of  the  reaction  at  960°  is,  therefore,  according  to 
our  method  of  Case  (6),— 38,560— 69,860  +  100,459  =  —7,961 


*  Approximate    determination    of    heat    in  Fe304,   made   recently  in 
author's  laboratory,  Q==  0.1447t  +  0,0001878t2. 


70  METALLURGICAL  CALCULATIONS. 

Calories.  It  thus  appears  that  Premier's  dilemma  was  mostly 
caused  by  his  thinking  that  the  thermochemical  value  of  the 
reaction  at  zero  should  be  a  constant  for  any  temperature:  a 
proper  thermochemical  calculation  removes  the  dilemma. 

Since  the  whole  treatment  of  the  subject  of  the  thermo- 
chemistry of  high  temperatures  requires  a  knowledge  of  data 
concerning  the  heat  capacity  of  elements  and  compounds  in  the 
solid,  liquid  and  gaseous  states,  as  well  as  of  their,  latent  heats 
of  fusion  and  vaporization,  the  next  instalment  of  these  cal- 
culations will  supply  these  data  as  far  as  they  are  known,  and 
discuss  a  number  of  applications  of  these  principles  to  various 
metallurgical  processes. 

In  order  to  apply  the  principles  explained  in  the  preceding 
discussion,  two  sets  of  data  are  necessary;  first,  the  thermo- 
chemical data,  such  as  are  ordinarily  obtained  by  laboratory 
experiments  at  laboratory  temperatures;  second,  physical  data 
concerning  the  specific  heats  and  latent  heats  of  fusion,  and 
volatilization  of  elements  and  compounds.  The  first  have 
been  given,  at  least  for  all  important  compounds  met  with  in 
metallurgy,  in  a  previous  place  (p.  18),  the  latter  will  now 
be  discussed  and  the  data  presented  as  far  as  they  have  been 
determined. 

SPECIFIC  HEATS  OF  THE  ELEMENTS. 

Dulong  and  Petit's  law  announces  the  fact  that  the  specific 
heat  of  atomic  weight  of  a  solid  metal  is  nearly  constant,  the 
value  varying  between  6  and  7,  and  averaging  6.4.  This  gen- 
eralization was  made  chiefly  upon  the  basis  of  the  specific  heats 
of  the  metals,  as  determined  in  the  range  100°  C.  to  10°  or  15° 
C.,  such  as  in  Regnault's  accurate  experiments.  About  the  only 
notable  exceptions  to  this  rule  are  carbon,  boron  and  silicon, 
and  it  has  been  naively  remarked  by  more  modern  physicists 
that  these  exceptions  to  the  rule  disappear  if  we  find  the  spe- 
cific heat  of  these  three  elements  at  high  temperatures,  that, 
for  instance,  the  specific  heat  of  carbon  above  1000°  C.  is  0.5. 
making  its  atomic  specific  heat  0.5X12  =  6,  and  therefore 
the  exceptions  to  the  rule  are  all  accounted  for.  Now,  the 
rule  is  an  important  one,  and  has  done  good  service,  but  the 
exceptions  just  noted  and  their  behavior  at  high  temperatures 
really  prove  that  the  rule  must  be  made  more  general,  or  else 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES..        71 

abandoned  altogether.  The  fact  is,  that  the  specific  heats  of 
almost  all  the  solid  elements  increase  with  the  temperature 
at  a  rate  equal  to  an  increase  of  about  0.04  per  cent,  of  their 
value  for  each  degree  centigrade,  so  that  the  atomic  specific 
heat  of  the  majority  of  the  elements,  which  is  about  6.4  at 
ordinary  temperatures,  becomes  about  40  per  cent,  greater  at 
1000°  C.  for  such  elements  as  are  not  melted  at  that  tempera- 
ture. Therefore,  while  we  may  say  that  at  ordinary  tem- 
peratures the  specific  heat  of  atomic  weight  of  a  solid  element 
is  6.4,  and  its  specific  heat  per  unit  of  weight  is  6.4,  divided 
by  the  atomic  weight,  yet  it  will  be  more  accurate,  if  actual 
determinations  have  not  been  made,  to  assume  that  the 
actual  specific  heat  increases  0.04  per  cent,  for  every  degree 
rise  in  temperature,  and  mean  specific  heat  to  zero  half  that 
fast. 

The  specific  heat  in  the  liquid  state  has  not  been  determined 
for  many  elements.  It  is  in  general  higher  than  the  specific 
heat  of  the  solid  at  ordinary  temperatures;  in  fact,  it  appears 
to  be  more  nearly  equal  to  the  specific  heat  of  the  solid  element 
just  before  fusion,  and  may  be  so  assumed  if  no  determina- 
tions have  been  made.  It  is  found,  furthermore,  not  to  change 
perceptibly  with  rise  of  temperature,  so  that  it  may  be  as- 
sumed constant. 

The  specific  heat  of  the  gaseous  elements  has  been  deter- 
mined only  for  those  which  are  gaseous  at  low  temperatures. 
For  the  metals  which,  as  far  as  known,  have  monatomic  vapors, 
i.e.,  vapors  in  which  the  atoms  exist  alone  and  uncoupled  with 
each  other,  the  specific  heat  of  atomic  weight,  occupying  22.22 
cubic  meters  at  standard  conditions,  should  be  theoretically  5.0 
Calories  at  constant  pressure,  or  0.225  per  cubic  meter.  We 
may  thus  estimate  the  specific  heat  of  metallic  vapors  which 
have  not  been  determined. 

LATENT  HEATS  OF  FUSION  OF  THE  ELEMENTS. 

The  passage  from  the  solid  to  the  liquid  state  is  in  all  cases 
accompanied  by  an  absorption  of  heat,  which  in  amount  varies 
from  one  or  two  Calories  up  to  100  Calories  per  unit  of  weight. 
This  quantity  has  been  most  frequently  determined  by  finding 
calorimetrically  how  much  heat  is  given  out  by  unit  weight 
of  the  melted  element  just  at  its  setting  point,  in  cooling  to 


72  METALLURGICAL  CALCULATIONS. 

ordinary  temperatures,  and  subtracting  from  this  the  heat  in 
unit  weight  of  the  solid  substance  at  the  melting  point,  ae 
determined  most  accurately  by  exterpolating  the  value  of 
mean  specific  heat  of  the  solid  up  to  the  melting  point.  In 
this  manner  the  latent  heat  of  fusion  for  a  number  of  elements 
has  been  directly  determined. 

If  a  crucible  full  of  melted  metal  is  allowed  to  cool,  the 
temperature  falls  regularly  until  the  melting  point  is  reached, 
and  then  stays  constant,  or  nearly  so,  for  some  time,  while 
the  metal  is  setting.  A  comparison  of  the  rate  of  cooling 
before  and  after  setting,  with  the  length  of  time  during 
which  the  temperature  was  constant,  gives  the  relative  value 
of  the  latent  heat  of  fusion  in  terms  of  the  specific  heat  of 
the  melted  metal  and  of  the  solid  metal  near  to  the  melting 
point. 

While  the  latent  heat  of  fusion  per  unit  weight  shows  no 
perceptible  regularities,  it  is  found  that  as  soon  as  the  latent 
heat  of  fusion  is  expressed  per  atomic  weight  of  the  element 
(analogous  to  specific  heat  of  atomic  weight)  that  notable 
regularities  appear.  The  elements  with  high  melting  points 
have  high  atomic  heats  of  fusion,  and  vice  versa,  so  that  if 
the  elements  are  arranged  in  the  order  of  their  melting  points 
their  latent  heats  of  fusion  per  atomic  weight  of  each  are  in 
the  same  order,  and  very  nearly  proportionately  so.  If,  for 
instance,  a  chart  or  diagram  is  made,  using  the  melting  points 
as  abscissas,  and  latent  heats  of  fusion  of  atomic  weights  as 
ordinates,  the  latter  will  lie  very  nearly  in  a  straight  line. 
Numerically,  if  the  melting  points  be  expressed  in  degrees  of 
absolute  temperature  (centigrade  temperatures  plus  273),  the 
latent  heats  of  fusion  of  atomic  weights  average  about  2.1 
times  the  temperature  of  the  melting  point.  This  rule  may  be 
used  to  predict  an  undetermined  latent  heat  of  fusion. 

In  addition  to  the  above  general  rule,  another  one  bearing  on 
the  same  question  was  also  discovered  and  applied  by  the 
writer.  (See  Journal  of  the  Franklin  Institute,  May,  1897.) 
According  to  this  observation,  the  continued  product  of  the 
latent  heat  of  fusion  of  atomic  weight  and  the  coefficient  of 
expansion  and  the  cube  root  of  the  atomic  volume  (atomic 
weight  divided  by  specific  gravity)  is  a  constant.  If  the  co- 
efficient of  linear  expansion  between  0°  and  100°  C.  is  used. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         73 

the  constant  is  0.095,  or  if  the  actual  linear  expansion  of  unit 
length  from  0°  to  100°  C.  is  used,  the  constant  is  9.5.  This 
rule,  applied  to  all  elements  whose  latent  heat  of  fusion  is 
known,  gives  satisfactory  agreements,  and  enables  us  therefore 
to  predict  the  latent  heat  of  fusion  of  nearly  a  dozen  other 
elements  for  which  the  coefficient  of  expansion  is  known. 

LATENT  HEATS  OF  VAPORIZATION  OF  THE  ELEMENTS. 

This  datum  has  been  determined  for  but  a  very  few  elements. 
Some  of  the  metalloids,  like  sulphur,  phosphorous  and  arsenic, 
are  known  to  become  complex  vapors  immediately  above  their 
boiling  point,  corresponding  to  such  formulae  as  S6,  P4,  As4; 
the  metals,  as  far  as  they  have  been  tested,  pass  into  monatomic 
vapors,  such  as  Na,  K,  Hg,  Zn  and  Cd,  in  which  each  atom 
represents  a  molecule.  In  the  latter  cases  the  following  gen- 
eralization may  be  made:  The  latent  heat  of  vaporization  of 
atomic  weight  is  proportional  to  the  absolute  temperatures  of 
the  boiling  point  at  atmospheric  pressure,  and  is  numerically 
equal  to  about  twenty-three  times  that  temperature  (twenty- 
one  times,  if  the  outer  work  of  overcoming  the  atmospheric 
pressure  be  not  included).  From  this  rule  it  is  possible  to 
estimate  the  amount  of  heat  necessary  to  vaporize  any  metal 
whose  boiling  point  under  atmospheric  pressure  is  known. 

Examples'.  The  boiling  point  of  carbon  under  atmospheric 
pressure  is  3,700°  C.,  and  if  its  vapor  is  monatomic,  the  latent 
heat  of  vaporization  is,  for  an  atomic  weight  of  carbon  (C  = 
12),  23X  (3700  +  273)  =  92,080  Calories,  equal  to.  7,673  Calories 
per  kilogram  of  carbon.  If  the  vapor  is  diatomic,  and  its  for- 
mula C2,  then  the  above  latent  heat  is  for  twenty-four  parts  of 
carbon,  and  for  one  part  by  weight  is  3,837  Calories.  Other 
considerations,  from  thermochemistry,  make  the  latter  value 
the  more  probable  one. 

The  boiling  point  of  cadmium  is  772°  C.,  and  its  vapor  is 
known  to  be  monatomic,  what  is  its  latent  heat  of  vaporiza- 
tion? The  atomic  weight  being  LI  2,  the  latent  heat  of  vapor- 
ization of  this  quantity  is  23  X  (772 +  273)  =  24,035  Calories, 
which  is  215  Calories  per  kilogram. 

In  making  such  calculations  it  must  be  strictly  observed  that 
the  boiling  point  under  atmospheric  pressure  is  to  be  used,  and 
not  any  temperature  at  which  vapors  may  appear  at  partial 


74  METALLURGICAL  CALCULATIONS. 

tensions  which  may  be  only  small  fractions  of  atmospheric 
pressure. 

THERMOPHYSICS  OF  THE  ELEMENTS. 

Having  laid  down  the  laws  and  the  empirical  rules  which 
appear  to  govern  these  phenomena,  we  will  now  discuss  the 
data  for  the  common  elements,  giving  both  what  is  known  and 
what  may  be  assumed  as  probably  true  wherever  actual  deter- 
minations have  not  been  made.  The  elements  will  be  taken 
up  in  the  order  of  their  atomic  weights, — the  only  scientifically 
logical  order. 

In  all  cases,  the  actually  measured  mean  specific  heats,  Sm, 
will  be  given.  In  the  case  of  gases,  these  will  be  the  mean 
specific  heats  under  constant  pressure ;  if  they  are  desired  under 
constant  volume  the  amount  of  outer  work  must  be  calculated 
in  Calories  (two  Calories  per  degree  for  a  molecular  weight 
of  a  gas,  0.09  Calories  per  degree  for  1  cubic  meter,  and  0.09^ 
weight  of  1  cubic  meter  for  a  kilogram  of  gas),  and  subtracted 
from  the  specific  heat  at  constant  pressure.  The  specific 
heat  of  1  cubic  foot  is,  in  pound  Calories,  the  specific  heat  per 
cubic  meter  divided  by  35.32  and  multiplied  by  2.204,  or,  in 
brief,  multiplied  by  0.0624;  in  British  thermal  units  it  will 
be  the  same  as  in  pound  Calories,  since  t  is  then  Fahrenheit 
degrees. 

If,  from  the  data  given,  it  is  desired  to  find  the  mean  specific 
heat  between  any  two  temperatures,  t  and  t',  instead  of  the 
mean  specific  heat  from  t  to  0°,  as  given  directly  by  the  for- 
mula, it  need  only  be  observed  that  Sm  from  t'  to  t  is  obtain- 
able by  finding  Sm  from  0°  to  (t'  +  t).  If,  for  instance,  Sm 
(o  to  t)  =  0.303 +  0.000027t,  then  Sm  (t  to  t')  =  0.303  + 
0.000027  (t'+t).  Furthermore,  if  the  actual  specific  heat  at 
any  temperature  t  is  desired,  it  is  equal  to  the  mean  specific 
heat  from  0°  to  2t;  e.g.,  in  the  above  cases,  S  (at  any  tempera- 
ture t)  =  0.303  +  0.000027  (2t)  =  0.303  +  0.000054  (t). 

Temperatures  will  be  always  given  and  represented  in  centi- 
grade degrees,  except  where  the  specific  heat  is  given  in  British 
thermal  units,  in  which  case  /  represents,  of  course,  Fahrenheit 
degrees,  and  will  also  be  printed  in  italics,  /,  to  further  dis- 
tinguish it  from  t  in  centigrade  degrees.  Absolute  tempera- 
tures, if  used,  will  be  designated  as  T,  and  are  equal  to  t  +  273. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         75 

The  vapor  tensions  of  all  the  elements  are  expressed  in  the 

^ 
form  log  p  =  —  ™+B,  where  p  is  used  in  millimeters  of  mercury,  T 

is  the  corresponding  temperature  in  absolute  measure  (K°),  and 
A  and  B  are  constants  for  the  elements  in  question,  which  can  be 
found  if  merely  two  values  of  p  and  T  are  known.  The  latent 
heat  of  vaporization  of  molecular  volume  is,  thermodynamically, 
4.57A.  B  is  found  nearly  constant  for  all  elements,  averaging 
7.9  to  8.4  in  the  case  of  the  liquid  element  and  8.85  for  the  solid 
element.  If  only  one  value  of  p  and  T  are  known,  then  B  is 
assumed  either  7.9  for  elements  of  low  boiling  point  or  8.4  for 
elements  of  high  boiling  point.  A  little  study  of  the  question 
will  show  that  these  values  correspond  to  using  23  or  25.1  for 
Trouton's  constant  (KT)  expressing  the  latent  heat  of  vapori- 
zation of  molecular  volume  at  p  =  760  mm.  (A/T  =  Trouton's 
constant/4.57).  The  vapor  tension  at  the  melting  point  is  cal- 
culated by  putting  T  =  M.  P.,  at  which  point  the  liquid  and  solid 
have  the  same  vapor  tension.  This  gives  one  value  of  p  and  T 

A' 
in  the  analogous  formula  for  the  solid  state,  (log.  p=  —  -~r+B'), 

where  A'  can  be  calculated  from  A  and  the  latent  heat  of  fusion, 
and  B'  thus  becomes  known.  The  vapor  tension  down  to  0°  C. 
is  then  calculated  from  this  formula. 

HYDROGEN. 

From  the  experiments  of  Mallard  and  LeChatelier  we  deduce : 
Sm  (0-t)   1  kilogram  (up  to  2000°  C.)    =3.370+0.0003t 

Calories. 
1  pound  (up  to  2000°  C.)        =3.370+0.0003t 

pound  Calories. 
1  pound  (up  to  3600°  F.)       =3.370+0.00017* 

B.  T.  U. 
1  cu.  meter  (up  to  2000°  C.)    =  0.303 +0.000027t 

Calories. 
1.  cu.  foot  (up  to  2000°  C.)      =0.0189+0.0000017t 

pound  Calories. 
1  cu.  foot  (up  to  3600°  F.)      =0.0189+0.0000009* 

B.  T.  U. 
For  higher  temperatures,  such  as  electric  furnace  heats  be- 


76 


METALLURGICAL  CALCULATIONS. 


tween  2000°  and  4000°  C.  (3600°  to  7200°  P.),  Berthelot  and 
Vielle  have  made  experiments  which  give  us : 


Sm  (0 — t)  1  kilogram 
1  pound 
pound 


=  2.75     +0.0008t  Calories. 

=  2.75     +0.0008t  pound  Cal. 
1  pound  =  2.75     +0.00044*  B.  T.  U. 

1  cubic  meter  =  0.2575 +0.000072t  Calories. 
1  cubic  foot     =  0.0161 +0.0000045t  pound  Cal. 
1  cubic  foot     =  0.0161+0.0000025*  B.  T.  U. 


LITHIUM. 


Sm  (0— t) 

S  (at  M.  P.  =  179°) 
Q  (solid,  at  M.  P.) 
L.  H.  Fusion 


Q  (liquid,  at  M.  P.) 

S  (liquid) 

Q  (liquid  at  B.  P.  76o  =  1450°) 

L.  H.  Vaporization  (1450°) 

Q  (vapor;  1450°) 
S  (vapor)  per  cubic  meter 
per  kilo 

Vapor  tension,  liquid  :        log  p 


0.7895+0.0024t+ 

0.0000063t2  (Bernini). 
2.255. 
254  Cal. 

136  Cal.  (calculated  by  2.1 
Trule). 
173  Cal.  (calculated  by 

second  rule) 
390  Cal.  (calculated). 
2.255  (assumed). 
3483  Cal.  (calculated). 
5660    Cal.    (Trouton's   rule, 
K  =  23). 

9140  Cal.  (calculated). 
0.225  (assumed). 
0.714  (calculated).  ' 

-^+7.92 


at  M.  P. 
solid : 
at  0°  C. 


logp  =  - 


(B  =  7.92,  assumed). 
4.9XlO-12mm.  (calculated). 

^2+8.38. 


=  6.0X10-25mm.  (calculated). 


BERYLLIUM  (GLUCINUM). 

Sm  (0-t)  =  0.38+0.0004t  (Nilson  &  Pettersson). 

S(atM.P.  =  1430°)  =  1.52. 
Q  (solid,  at  M.  P.)      =  1358  Cal. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        77 

L.  H.  Fusion  ,     =  400  cal.  (calculated  by  2.1  T  rule). 

Q  (liquid,  at  M.  P.)  =  1758  Cal. 

S  (liquid)  =1.52  (assumed). 

BORON. 

Sm  (0— t)  (t  under  233°)          =0.22  +  0.00035t  (Weber). 
S  (solid,  at  500°)  =  0.57  (by  extension  of  above 

formula). 
S  (solid,  above  500°)  =  0.57   (assumed  to  remain 

constant) .  * 

88 
Sm  (0— t)  (t  over  500°)  =  0.57  — ' 

Q  (solid,  at  M.  P.  =  2500°)    =  1337  Cal.  (calculated). 

L.  H.  Fusion  =  265  Cal.    (calculated  by  2.1 

T  rule,  and  molecule  B2). 

Q  (liquid,  at  M.  P.)  =1602  Cal   (calculated). 

S  (liquid)  =  0.57  (assumed). 

Q  (liquid  at  B.  P.76o  =  3700°)  =  2285  Cal. 
L.  H.  Vaporization  (3700°)      =  4155  Cal.   (Trouton's  rule, 

K  =  23,  and  molecule  B2). 
Q  (vapor,  3700°)  =  6440  Cal. 

S  (vapor)  per  cubic  meter       =  0.315  (assumed  for  B2). 
per  kilo  =  0.318  (calculated). 

Vapor  tension,  liquid :     log  p  = ^ |-7.92  (B  =  7.92 

assumed), 
at  M.  P.  =5  mm. 

solid:      logp=  -?M?9+8.36. 

at  0°  C.  =  3.9X10-70  mm.   (calculated). 

CARBON. 

{Amorphous  and  Graphitic.) 
Sm  (0°— 1°)  (up  to  250°)  =  0.1567+0.00036t. 

(0°—t°)  (250°— 1000°)        =  0.2142  +  0.000166t. 
S  at  1000°  =  0.528  (Violle);  0.546  (Weber). 

Q  at  1000°  =  380  Cal. 

Sm  (0-t)  (t  over  1000°)  =  0.4430 +0.0000425t-^^ 

(Violle). 

*This  figure  would  give  atomic  specific  heat  =  6.27,  and  we  assume  this 
constancy  by  analogy  with  similar  behavior  of  carbon. 


78  METALLURGICAL  CALCULATIONS. 

Q  (0—  t)  (t  over  1000°)  =  0.4430t+0.0000425t2-  106.5. 

Q  (solid)  (atB.  P.76o  =  3700°)  =  2114  Cal. 

L.  H.  Sublimation  (3700°)        =  4140  Cal.  (Richard's  rule, 

K  =  2.1,  and  Trouton's  rule,  K  =  23;forC2), 
Q  (vapor)  (3700°)  =  6254. 

S  (solid,  at  M.  P.  =  4400°)      =  0.777  (Violle's  equation). 
S  (vapor)  per  cubic  meter         =  0.315  (assumed). 

per  kilo  =  0.292  (calculated  for  C2). 

Q  (solid)  (at  M.  P.  4400°)         =  2666  Cal.  (by  Violle's  equation). 
L.  H.  Fusion  (4400°)  =  409  Cal.  (by  2.1  T  rule  for  C2). 

=  396  Cal.  (by  second  rule,  for  C2). 

Q  (liquid)  (4400°)  =  3066  Cal.  (calculated). 

S  (liquid)  =  0.777  (assumed). 

Q  (vapor)  (4400°)  =  6806  Cal.  (calculated). 


24 
Vapor  tension,  solid:        log  p  =  -^jr+9.0  (B'  =  9.0, 

assumed)  . 

at  M.  P.  =  6310  mm.  =  8.3  atmospheres. 

r     -A       1  22,195 

liquid:     log  p  =  --  ^  --  1-8.5. 

at  0°C.  =  1  X  1C-80  mm.  (calculated). 

NITROGEN. 

Sm  (0—  t)  1  kilogram  (up  to  2000°  C.)     =  0.2405  +0.0000214t 

Cal. 
1  pound  (up  to  2000°  C.)         =  0.2405+0.  00002  14t 

pound  Cal. 
1  pound  (up  to  3600°  F.)          =  0.2405+0.0000119* 

B.  T.  U. 
1  cubic  meter  (up  to  2000°  C.)  =  0.303  +0.000027t 

Cal. 
1  cu.  foot  (up  to  2000°  C.)         =  0.0189  +0.0000017t 

pound  Cal. 

1  cu.  foot  (up  to  3600°  F.)        =  0.0189+0.0000009* 

B.  T.  U. 

For  temperatures  between  2000°  and  4000°  C.  the  following 
are  the  values  of  the  mean  specific  heats  to  zero  : 

Sm  (0—  t)  1  kilogram        =  0.2044  +0.000057t  Cal. 

1  pound  =  0.2044+0.000057t  pound  Cal. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        79 

1  pound  =  0.2044+0.000032*  B.  T.  U. 

1  cubic  meter  =  0.2575 +0.000072t  Cal. 

1  cubic  foot  =  0.1601 +0.0000045t  pound  Cal. 

1  cubic  foot  =  0.0161+0.0000025*  B.  T.  U. 

OXYGEN. 

Sm  (0— t)  1  kilogram  (up  to  2000°  C.)  =  0.2104+0.0000187t 

Cal. 
1  pound  (up  to  2000°  C.)       =  0.2104+0.0000187t 

pound  Cal. 
1  pound  (up  to  3600°  F.)       =  0.2104+0.0000104* 

B.  T.  U. 

1  cu.  meter  (up  to  2000°  C.)  =  0.303 +0.000027t  Cal. 
1  cu.  foot  (up  to  2000°  C.)     =  0.0189+0.0000017t 

pound  Cal. 
1  cu.  foot  (up  to  3600°  F.)     =  0.0189+0.0000009* 

B.  T.  U. 

Sm  (0— t)  1  cu.  meter  (2000°  -4000°C)  =0.2575 +0.000072t  Cal. 
1  cu.  foot  (2000°- 4000°  C.)  =  0.0161 +0.0000045t 

pound  Cal. 
leu.  foot  (3600° -7200°  F.)  =  0.0161+0.0000025* 

B.  T.  U. 

1  kilogram  (2000°-4000°  C)  =  0.1788+0.00005t  Cal. 
1  pound  (2000° -4000°  C.)  =  0.1788+0.00005t 

pound  Cal. 
1  pound  (3600°-7200°  F.)  =  0. 1788+0. 00003 1 

B.  T.  U. 

SODIUM. 

Sm  (0— t)  =  0.2932+0.00019t  (Bernini). 

Q  (solid,  at  M.  P.  =  98°)  =  30.6  Cal.  (by  formula). 

L.  H.  Fusion  =  27.2  (Rengade). 

Q  (liquid,  at  M.  P.)  =  57.8  Cal. 

S  (liquid,  98°— 100°)  =  0.333  (Bernini). 

Q  (liquid  at  B.  P.760  =  877°)  =  317  Cal.  (calculated). 

L.  H.  Vaporization  (877°)  =  905  Cal.  (from  vapor  tension 

constants) . 

Q  (vapor,  at  877°)  =  1222  Cal. 

S  (vapor)  for  cubic  meter  =  0.225  (assumed), 

per  kilo  =  0.218  (calculated). 


80  METALLURGICAL  CALCULATIONS. 

Vapor  tension,  liquid:    logp  =  --  ^-+6.85  (from  Gebhardt's 

data) 

at  M.  P.  =  3.55  X10~6  mm.  (calculated). 

4700 
solid:     log.p  =-  -4jF+7.30. 

at  0°  C.  =  8.1  X  10~9  mm.  (calculated). 

MAGNESIUM. 

Sm  (0—  1°)  =  0.2372+0.000093t+ 

0.0000000685t2  (Stucker). 

Q  (solid,  at  M.  P.  =  650°)         =  212  Cal.  (by  formula). 
L.  H.  Fusion  =    70  Cal.  (Roos). 

Q  (liquid,  at  M.  P.)  =  282  Cal. 

S  (liquid)  =  0.445  (assumed;  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.76o  =  1120°)  =  491  Cal.  (calculated). 
L.  H.  Vaporization  (1120°)         =  1443     Cal.     (Trouton's    rule, 

K  =  25.2). 

Q  (vapor,  at  1120°)  =  1934  Cal.      . 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

kilo  =  0.208  (calculated). 

Vapor  tension,  liquid  :  log  p       =  --  ~-+8.4  (constants  from 

Trouton's  rule). 
at  M.  P.  =1.15  mm.  (calculated). 


solid  :  log  p      =  -          +  8-8L 

at  0°  C.  =  5.25  X10~20  mm.  (calculated). 

ALUMINIUM. 

Sm(0—  1°)  =  0.2220+0.00005t  (Richards). 

Q  (solid,  at  M.  P.    =  657°)  =*=  167.4  Cal.  (from  formula). 

L.  H.  Fusion  =  90.9  Cal.  (Richards). 

Q  (liquid,  at  M.  P.)  =  258.3  Cal.  (Richards). 

S  (liquid)  =  0.308  (Pionchon). 

Q  (liquid,  at  B.  P.76o  =  2200°)  =  733.5  Cal.  (calculated). 

L.  H.  Vaporization  (2200°)  =  2305    Cal.     (Trouton's    rule, 

K  =  25.2). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        81 

Q  (vapor,  at  2200°)  =  3039  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

kilo  =  0.185  (calculated). 


i  Q 

Vapor  tension,  liquid  :      log  p  =  --  '^  --  h8.4  (constants  from 

Trouton's  rule). 
at  M.  P.  =  5.25  X10~7  mm.  (calculated). 

14  1QO 
solid  :       log  p  =  -^^+8.98. 

at  0°  C.  =  1.0X10-43  mm.  (calculated). 

SILICON. 

Sm(0—  t°)(up  to  234°)  =  0.17+0.00007t  (Weber). 

Q  (solid,  at  M.  P.  =  1430°)       =  386  Cal.  (by  extension  of  for- 

mula). 
L.  H.  Fusion  =  128  Cal.  (by  2.1  T  rule). 

=  108  Cal.  (by  second  rule). 
Q  (liquid,  at  1430°)  =  494  Cal. 

S  (liquid)  =  0.37  (assumed,  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.760  =  2300°)  =  816  Cal.  calculated). 
L.  H.  Vaporization  (2300°)         =  1153  Cal.  (Trouton's  rule,  K 

=  25.2,  vapor  =  Si2). 
Q  (vapor,  at  2300°)  =  1969  Cal. 

S  (vapor)  per  cubic  meter          =  0.315  (assumed,  for  812). 
per  kilo  =  0.125  (calculated). 

14  200 

Vapor  tension,  liquid  :      log  p  =  --  ^  --  1-8.4  (constants  from 

Trouton's  rule). 
at  M.  P.  =  1.32  mm.  (calculated). 

solid:        logp=  - 

at  0°  C.  =  8X  10~47  mm.  (calculated). 

PHOSPHORUS. 

Sm(-20to7°)  ,=  0.1788. 

Q  (solid,  at  M.  P.  =  44°)  =  8  Cal. 

L.  H.  Fusion  (44°)  =  5  Cal.  (Person). 

Q  (liquid,  at  44°)  =  13  Cal. 

S  (liquid,  44°  to  98°)  =  0.205. 

Q  (liquid  at  B.  P.76o  =  287°)      =  63  Cal. 


82 


METALLURGICAL  CALCULATIONS. 


L.  H.  Vaporization  (287°) 
Q  (vapor,  at  287°) 
S  (vapor)  per  cubic  meter 
per  kilo 


=  130  Cal. 

=  193  Cal. 

=  0.405  (assumed,  for  P3). 

=  0.097  (calculated). 


Vapor  Tension,  liquid  :     log  p  =  — 
at  M.  P. 
solid  : 
at  0°  C. 


2160 


+7.0. 


T 
=  1.63  mm. 

logp=_?26C>+7.34. 


T 
=  0.115  mm. 


SULFUR. 


Sm(15°— 97°) 

Q(solid,  at  M.  P.  =  113°) 

L.  H.  Fusion 

Q  (liquid,  at  M.  P.) 

Sm(0—t°)  (liquid,  114°— 445°) 

Q  (liquid  at  B.  P.76o  =  444°.5) 
L.  H.  Vaporization  (444°.5) 
Q  (vapor,  at  444°.5) 
S  (vapor)  per  cubic  meter 
per  cubic  meter 
per  cubic  meter 
per  kilo 


Vapor  tension,  liquid  :      log  p 
at  M.  P. 
solid  :        log  p 
atO°C. 


0.18  (Regnault). 

20  Cal. 

9.4  (Person). 

29.4  Cal. 

0.338+0.000187t- 

0.00000058t2. 
126  Cal. 
72  Cal. 
198  Cal. 

0.855  for  S8,  close  to  445°. 
0.674  for  S6,  up  to  500°. 
0.315  for  S2,  over  800°. 
0.074  for  S8,  close  to  445°. 
0.078  for  S6,  up  to  500°. 
0.109  for  S3,  over  800°. 

-^+8.1. 

0.028  mm. 
4140 


T 
lXlO-6mm. 


CHLORINE. 


Sm  (gas)  per  cubic  meter 
per  kilo 


=  0.40  (13°— 202°)  (Regnault). 
=  0.1241  (Regnault). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         83 
POTASSIUM. 

Sm(0— 1°)  =  0.1858+0.00008t  (Bernini). 

Q  (solid,  at  M.  P.    =  60°)  =  11.4  Cal. 

L.  H.  Fusion  =  13.6  Cal.  (Bernini). 

Q  (liquid,  at  M.  P.)  =  25.0  Cal. 

Sm  (liquid)  (0— 1°)  =  0.1422+0.00067t  (Rengade). 

Q  (liquid,  at  B.  P.760  =  757°)  =  181.5  Cal. 

L.  H.  Vaporization  (757°)  =  607  Cal.  (Trouton's  rule, 

K  =  23). 

Q  (vapor,  at  757°)  =  789  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed), 

per  kilo  =  0.128  (calculated). 

Vapor  Tension,  liquid  :    log  p  = jp— +7.92  (constants  from 

-^  Trouton's  rule), 

at  M.  P.  =  2.25  X10~8  mm. 

solid  :        logp  =  - 
at  0°C.  =  7.lXlO-12mm. 

CALCIUM. 

Sm(0°— 100°)  =  0.1704  (Bunsen). 

Q  (solid,  at  M.  P.  =  800°)         =  136.3  Cal. 
L.  H.  Fusion  =  56.3  Cal.   (calculated   by    2.1 

T  rule). 
Q  (liquid,  at  M.  P.)  =  192.6. 

TITANIUM. 

S  (solid,  at  0°)  =  0.0978   (Nilson  and 

Pettersson) . 

Sm  (0— 1°)  =  0.0978+0.0000035t       (assum- 

ing atomic  specific  heat  =  10 
at  M.  P.). 

Q  (solid,  at  M.  P.  =  1795°)          =  187  Cal.  (from  above  formula). 
L.  H.  Fusion  =  90  Cal.  (from  2.1  T  rule). 

Q  (liquid,  at  M.  P.)  =  277  Cal. 

S  (liquid)  =  0.2084     (assumed,     same     as 

solid  at  M.  P.). 
Q  (liquid,  at  B.  P.76o  =  2475°)  =  319  Cal. 


84  METALLURGICAL  CALCULATIONS. 

L.  H.  Vaporization  (2475°)         =  143  Cal.  (from  Trouton's  rule, 

K  =  25). 

Q  (vapor,  at  2475°)  =  462  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.104  (calculated). 

Vapor  tension,  liquid  :      log  p  =  --  ^  --  \-  8.4. 
at  M.  P.  =  11.75  mm. 

solid  :      log  p  =  --  ^p?  +  8.85> 
atO°C.  =  8.3XlO-49mm. 

VANADIUM. 

Sm  (0—  1°)  =  0.105+0.00003t. 

Q  (solid,  at  M.  P.  =  1710°)       =  267  Cal. 
L.  H.  Fusion  =  82  Cal.  (by  2.1  T  rule). 

Q  (liquid,  at  M.  P.)     ,  =  349  Cal. 

S  (liquid)  =  0.208  (assumed,  same  as  solid 

at  M.  P.). 

CHROMIUM. 

Sm  (0—  1°)  (to  600°)  =  0.1039+0.00000008t2  (Adler). 

Q  (solid,  at  M.  P.  =  1489°)       =  352  Cal. 

L.  H.  Fusion  =  71  Cal.  (calculated  by  2.1  T 

(rule). 

Q  (liquid,  at  M.  P.)  =  423  Cal. 

S  (liquid)  =0.24  (estimated). 

Q  (liquid  at  B.  P.76o  =  2200°)    =  594  Cal. 
L.  H.  Vaporization  (at  2200°)    =  1197   Cal.   (Trouton's  rule, 

K   =  25.1). 

Q  (vapor,  at  2200°)  =  1791  Cal. 

S  (vapor),  per  cubic  meter          =  0.225  (assumed). 
per  kilo  =  0.096 


1  o 

Vapor  tension,  liquid  :     log  p.  =  --  ^  --  h  8.  4  (const  ants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  4.47  mm. 


14 
solid  :     log  p.  =  -  --  +  8.86. 

at  0°  C.  =  7.8  X  10~45  mm. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        85 

MANGANESE. 

Sm  (0— 1°)  =  0.1088+0.000103t. 

Q  (solid,  at  M.  P.  =  1207°)       =  281  Cal. 

L.  H.  Fusion  =  43  Cal.  (calculated  by  second 

rule.) 

Q  (liquid,  at  M.  P.)  =  324  Cal. 

S  (liquid)  =  0.357  (assumed  from  S  solid  at 

M.  P.). 

Q  (liquid  at  B.  P.76o  =  1900°)    =  572  Cal. 
L.  H.  Vaporization  (1900°)         =  995  Cal.  (Trouton's  rule, 

K  =  25.1). 

Q  (vapor,  at  1900°)  =  1567  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed), 
per  kilo.  =  0.091  (calculated). 

11  995 

Vapor  tension,  liquid  :      log  p  = ^ h  8.4    (Constants 

from  Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  1.98  mm.  Hg. 

solid  :     log  p.  =  -  i^P  +  8.85. 
at  0°  C.  =  3.2X10-38mm.  Hg. 

IRON. 
Sm(0— 1°)  up  to  660°  =  0.1 1012+0. 000025t+ 

0.0000000547t2    (Pionchon; 
also  Oberhoffer). 

Q  at  660°  =  96.1  Cal. 

Q  at  750°  =  125.6  Cal. 

L.H.  Change  of  state  (730°)        =5.3    Cal.    (Pionchon);    5.0 

(Leschtschenko) . 

S  above  750°  =  0.1675     (approximately     con- 

stant.) 

Q  (0— 1°)  t  =  750°  to  1500°       =  0.1675t  -  51  Cal. 
L.  H.  Change  of  state  (900°)      =  6.0  Cal.  (Pionchon);  6.1 

(Leschtschenko) . 

Q  (solid,  at  M.  P.    =  1535°)      =  256  Cal. 
L.  H.  Fusion  =  66  Cal.  (calculated  by  2.1  T 

rule). 

=  69  Cal.  (calculated  by  second 

rule). 
(4.3%  C.)  =  59  Cal.  (Schmidt). 


86  METALLURGICAL  CALCULATIONS. 

Q  (liquid,  at  1535°)  =  322  Cal. 

S  (liquid)  =  0.20  (estimated). 

Q  (liquid  at  B.  P.760  =  2450°)  =  505  Cal. 

L.  H.  Vaporization  (2450°)  =  1224  Cal.   (Trouton's  rule, 

K  =  25.1). 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo.  =  0.089  (calculated). 

Vapor  tension,  liquid  :     log  p.  =  —  —  ^  --  h  8.4       (Constants 

from  Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  1.20  mm.  Hg. 

solid:     log  p.  =  ~ 


.. 
=  at  0°  C.         =  6.9X10-50mm. 

NICKEL. 
Sm  (0—  1°)  up  to  230°  =  0.  10836  +0.00002233t 

(Pionchon)  . 
L.  H.  Change  of  State  =  4.64  Cal.,  230°  to  400°, 

(Pionchon)  . 
=  3.1.1  Cal.  363°  (Leschtschenko)  . 

Sm  (0—  1°)  t  =  440°  to  1050°     =  0.099  +0.00003375t  +  ^ 

t 

(Pionchon)  . 

Q  (solid,  at  M.  P.  =  1450°)       =  221  Cal. 

L.  H.  Fusion  =  68  Cal.  (calculated  by  second 

rule). 

Q  (liquid,  at  1450°)  =  289  Cal. 

S  (liquid)  =  0.197  (estimated). 

Q  (liquid,  at  B.  P.76o  =  2150°)   =  427  Cal. 
L.  H.  Vaporization  (2150°)         =  1042  Cal.  (Trouton's  rule, 

K=  25.1). 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.086  (calculated). 

13  375 

Vapor  tension,  liquid  :     log  p.  =  --  ^  --  f-  8.4    (Constants 

from  Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  4.36  mm.  Hg. 


solid  :     log  p.  =  -        p-  +  8.9. 
at  0°  C.  =  5.25  X  10~44  mm.  Hg. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         87 
COBALT. 

Sm  (0—  1°)  up  to  900°  =  0.105844-0.00002287t-f 

0.000000022t2  (Pionchon). 

Sm  (0—  t°)t  over  900°  =  0.124+0.00004t  -  ^ 


(Pionchon). 

Q  (solid,  at  M.  P.  =  1490°)         =  259  Cal.   (extension  of  above 

formula)  . 

L.  H.  Fusion  =    68  Cal.  (calculated  by  second 

rule). 

Q  (liquid,  at  1490°)  =  327  Cal. 

S  (liquid)  =  0.243  (assumed,  same  as  solid 

at  M.  P.). 

COPPER. 

Sm  (0—  t)  =  0.0939+0.00001778t    (Frazier 

and  Richards). 

Q  (solid,  at  M.  P.  =  1083°)       =  118.7  Cal. 
L.  H.  Fusion  =  43.3  Cal.  (Richards). 

Q  (liquid,  at  M.  P.)  =  162.0  Cal.  (Frazier  &  Richards)  . 

S  (liquid)  =  0.156  (Glaser). 

Q  (liquid,  at  B.  P.760  =  2310°)  =  353  Cal. 

L.  H.  Vaporization  (2310°)         =  1087  Cal.  (calculated,  from  va- 

por tension  curve). 

Q  (vapor,  at  2310°)  =  1440  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed). 
per  kilo  =  0.079  (calculated). 

Vapor  tension,  liquid  :     log  p.  =  --  ^  --  h  8.75  (from  Green- 

wood's data). 
at  M.  P.  =  3.5XlO-3mm.  Hg. 


solid  :    log  p.    =  -  +  9.17. 

at  0°  C.  =  2.6  X10~49  mm.  Hg. 

ZINC. 

Sm  (0—  1°)  =  0.0906  +0.000044t. 

Q  (solid,  at  M.  P.  =  419°)         =  45.2  Cal. 


88 


METALLURGICAL  CALCULATIONS. 


L.  H.  Fusion 

Q  (liquid,  at  M.  P.) 

S  (liquid) 

Q  (liquid,  at  B.  P.760  =  930°) 

L.  H.  Vaporization  (930°) 

-4 

Q  (vapor  at  930°) 
S  (vapor)  per  cubic  meter 
per  kilo 

Vapor  tension,  liquid  :      log  p  =  — 


at  M.  P. 
solid  :        log  p. 
at  0°  C. 


22.6  Cal.  (Richards). 
67.8  Cal. 
0.179  (Glaser). 
159  Cal. 

446  Cal.  (calculated  from  vapor 
tension  curve). 
605  Cal. 

0.225  (assumed). 
0.077  (calculated). 

/»O£>  pr 

—  ™  —  h  8.17  (from  various 
<     observations). 
9.3XlO-2mm.  Hg. 


=  1.03XlO-16mm.  Hg. 


Sm  (12°  to  23°) 

Q  (solid,  at  M.  P.  =  30°) 

L.  H.  Fusion 

Q  (liquid,  at  M.  P.) 

S  (liquid,  30°— 119°) 


Sm  (21°  to  65°)  amorphous 
crystalline 


Q  (solid,  at  S.  P.76o  =  616°) 
L.  H.  Sublimation  (616°) 

-» 

Q  (vapor,  at  616°) 
S  (vapor)  per  cubic  meter 

per  kilo 

Q  solid,  at  M.  P.  =  827°) 
L.  H.  Fusion  (827°) 
Q  (liquid  at  827°) 
L.  H.  Vaporization,  827°) 
Q  (vapor,  at  827°) 


GALLIUM. 

=  0.079  (Berthelot). 

=  2.4  Cal. 

=  19.3  Cal.  (Berthelot). 

=  21.7  Cal. 

=  0.084  (Berthelot). 

ARSENIC. 

=  0.0758  (Bettendorf  and 

Wullner). 

=  0.083  (Bettendorf  and 

Wullner). 

=  47  Cal. 

=  114  Cal.  (calculated  from  vapor 
tension  curve,  for  As3) . 

=  161  Cal. 

=  0.405  (assumed  for  Asa). 

=  0.040  (calculated). 

=    63  Cal/ 


=  10  Cal. 
=  73  Cal. 
=  104  Cal. 
=  177  Cal. 


under    10,000   mm. 
pressure. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        89 

Vapor  tension,  solid  :       log  p.  = jp-  +  9.10    (from  experi- 
mental data), 
at  M.  P.              =  10,000  mm.  Hg. 

liquid  :     log  p.  = ™ — h  8.65    (from  experi- 
mental data.) 
at  0°  C.               =  2.8  X10-21  mm.  Hg. 

SELENIUM. 

Sm  (60°— 200°)  =  0.084  (Bettendorf  and  Wullner). 

Q  (solid,  at  M.  P.  =  217°)      =  18  Cal. 
L.  H.  Fusion  =4  Cal. 

Q  (liquid  at  M.  P.)  =  22  Cal. 

S  (liquid)  =  0.12  (assumed). 

Q  (liquid,  at  B.  P,760  =  690°)  =  79  Cal. 

L.  H.  Vaporization  (690°)       =  97  Cal.  (calculated  from  vapor 

tension  curve,  for  Sea). 
Q  (vapor,  at  690°)  =  176  Cal. 

S  (vapor)  per  cubic  meter       =  0.405  (assumed  for  Sea), 
per  kilo  =  0.038  (calculated). 

Vapor  tension,  liquid  :  log  p  = ^ — f-8.10  (from  experi- 
mental data), 
at  M.  P.           =  0.71  mm.  Hg. 

solid  :  log.  p  = ™ — f-8.56 

atO°  C.  =  2.lXlO-11mm.Hg. 

BROMINE. 

Q  (solid,  at  M.  P.  =  -7°)     =  -16.9  Cal. 

L.  H.  Fusion  ( - 7°)  =  16.2  Cal.  (Regnault). 

Q  (liquid,  at  M.  P.)  =  -0.7  Cal. 

Sm  (liquid,  -6°  to  +58°)       =  0.105+O.OOllt 

Q  (liquid,  at  B.  P.76o  =  58°)  =  10  Cal. 

L.  H.  Vaporization  (58°)       =  43.7  Cal.  (Berthelot  and  Ogier). 

Q  (vapor,  at  58°)  =  53.7  Cal. 

S  (vapor)  per  cubic  meter     =0.40 

per  kilo  =  0.0555  (Regnault). 

1  A^?  f\ 

Vapor  tension,  liquid  :  log  p  = ™— +7.8    (from   experimental 

data). 

at  M.  P.        =  45  mm.  Hg. 
at  0°  C.         =  66  mm.  Hg. 


90  METALLURGICAL  CALCULATIONS. 

RUBIDIUM. 

Sm  (20°  to  35°)  =  0.0792+O.OOOOlt  (Deusz). 

Q  (solid,  at  M.  P.  =  38°)         =  3  Cal. 
L.  H.  Fusion  -  6  Cal.  (E.  Duess). 

Q  (liquid,  at  38°)  =  9  Cal. 

S  (liquid)  =  0.080  (assumed). 

Q  (liquid,  at  B.P.76Q  =  696°)  =  62  Cal. 

L.  H.  Vaporization  (696°)       =  261  Cal.   (from  Trouton's  rule; 

K  =  23). 

Q  (vapor,  at  696°)  =  323  Cal. 

S  (vapor)  per  cubic  meter       =  0.225  (assumed), 
per  kilo  =  0.059  (calculated). 

4890 
Vapor  tension,  liquid  :  log.  p  = ™— +7.90  (constants  from 

Trouton's  rule;  K  =  23). 
at  M.  P.         =  1.5 X10-8  mm.  Hg. 

solid  :     logp  = ~r~ +8.28. 

at  0°  C.  =  6.7  X  10~11  mm.  Hg. 

STRONTIUM. 

Sm  (0°— 100°)  =  0.0735  (assumed  by  Dulong  and 

Petit's  Law). 

Q  (solid,  at  M.  P.  =  800°)      =  59  Cal. 

L.  H.  Fusion  =  26  Cal.  (calculated  by  2.1  T  rule). 

Q  (liquid,  at  800°)  =  85  Cal. 

S  (liquid)  =  0.092  (assumed,  from  at.  sp. 

(heat  =  8.0). 

ZIRCONIUM. 

Sm  (0°— 100°)  =  0.0662  (Mixter  and  Dana) . 

Q  (solid,  at  M.  P.  =  2350°)    =  155  Cal. 

L.  H.  Fusion  =  61  Cal.  (calculated  by  2.1  T  rule). 

Q  (liquid,  at  2350°)  =  216  Cal. 

S  (liquid)  =  0.11  (assumed,  from  at.  sp. 

heat  =  10). 

COLUMBIUM  (NIOBIUM). 

Sm  (0—100°)  =  0.068    (assumed,    from    Dulong 

Q  (solid,  at  M.  P.  =  1950°)  =  133  Cal.  and  Petit's  Law). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        91 

L.  H.  Fusion  =  50  Cal.  (calculated  by  2.1  T  rule). 

Q  (liquid,  at  1950°)  =  183  Cal. 

S  (liquid)  =  0.107  (assumed    from  at.  sp. 

heat  =  10). 

MOLYBDENUM. 

Sm  (0°— 1°)  =  0.0655 +0.00002t  (Defacqz  and 

(Guichard). 

Q  (solid,  at  M.  P.  =  2500°)    =  289  Cal. 

L.  H.  Fusion  =  61  Cal.  (calculated  by  2.1  T  rule). 

Q  (liquid,  at  2500°)  =  350  Cal. 

S  (liquid)  =  0.165  (assumed,  from  extension 

of  formula  for  S  (solid) . 
Q(liquidatB.  P.760  =  3600°)  =  522  Cal. 
L.  H.  Vaporization  (3600°)      =  1015  Cal.  (calculated  from 

Trouton's  rule;  K  =  25). 
Q  (vapor,  at  3600°)  =  1537  Cal. 

S  (vapor)  per  cubic  meter       =  0.225  (assumed), 
per  kilo  =  0.052  (calculated). 

21  380 
Vapor  tension,  liquid  :  log  p  = ^ 1-8.4  (constants  from 

Trouton's  rule,  K  =  25). 
at  M.  P.          =  4.9  mm.  Hg. 

solid  :   log  p  =  -^j^+8.85. 
at  0°  C.  =  7.6  X  10~75  mm.  Hg. 

RUTHENIUM. 

Sm  (0°— 100°)  =  0.0611  (Bunsen). 

Sm  (0°— 1°)  =  0.0601 -f-O.OOOOlt  (function  of  t 

assumed). 

Q  (solid,  at  M.  P.=  1950°)     =  155    Cal.    (extension    of    above 

formula) . 

L.  H.  Fusion  =  46   Cal.    (calculated  by   second 

rule). 

Q  (liquid,  at  M.  P.)  =  201  Cal. 

S  (liquid)  =  0.099    (assumed   same  as   solid 

at  M.  P.). 

Q  (liquid  at  B.  P.76o  =  2520°)  =  257  Cal. 
L.  H.  Vaporization  =  689  Cal.  (calculated  from 

Trouton's  rule,  K  =  25). 


92  METALLURGICAL  CALCULATIONS. 

Q  (vapor,  at  2520°)  =  946  Cal. 

S  (vapor)  per  cubic  meter       =  0.225  (assumed), 
per  kilo  =  0.049  (calculated). 

Vapor  tension,  liquid  :  log  p    = ^ — (-8.4  (constants  from 

Trouton's  rule;  K  =  25). 
at  M.  P.  =  28.8  mm.  Hg. 

solid:     logp=  -  ^^+8.86. 
at  0°  C.  =  4X10-52  mm.  Hg. 

RHODIUM. 

Sm  (10°— 97°)  =  0.0580  (Regnault). 

Sm  (0°— 1°)  =  0.0574 +0.00001 1  (function  of  t 

assumed). 

Q  (solid,  at  M.  P.  =  1970°)    =  152  Cal. 

L.  H.  Fusion  --=  53   Cal.    (calculated  by  second 

rule). 

Q  (liquid,  at  M.  P.)  =  205  Cal. 

S  (liquid)  =  0.097  (assumed  same  as  solid  at 

M.  P. 

Q  (liquid at  B.  P.76o  =  2500°)  =  256  Cal. 
L.  H.  Vaporization  (2500°)     =  677  Cal.   (calculated  from 

Trouton's  rule,  K  =  25). 
Q  (vapor,  at  2500°)  =  933  Cal. 

S  (vapor)  per  cubic  meter       =  0.225  (assumed), 
per  kilo  =  0.049  (calculated). 

Vapor  tension,  liquid  :  log  p  = ^ f-8.4. 

at  M.  P.          =  38  mm.  Hg. 
solid:    logp=  -^P+8.94. 
at  0°  =  3.2  X  10~52  mm.  Hg. 

PALLADIUM. 

Sm  (0°— 1°)  =  0.0582+O.OOOOlt  (Violle). 

Q  (solid,  at  M.  P.  =  1549°)     =  110  Cal.  (Violle). 
L.  H.  Fusion  =  36  Cal.  (Violle). 

Q  (liquid,  at  M.  P.)  =  146  Cal.  (Violle). 

S  (liquid)  =  0.089  (assumed  same  as  solid  at 

M.  P.). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        93 

Q  (liquid,  atB.  P.76o  =  2535°)  =  238  Cal. 

L.  H.  Vaporization  (2535°)       =  667  Cal.  (calculated  from  Trou- 

ton's  rule;  K  =  25). 

Q  (vapor,  at  2535°)  =  905  Cal. 

S  (vapor)  per  cubic  meter         =  0.225  (assumed). 
per  kilo  =  0.047  (calculated). 

Vapor  tension,  liquid  :     log  p  =  --  ^  --  h8.4  (constants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  7.8  X  10-1  mm.  Hg. 

solid  :      log  p  =  -  ^i2  +  8>85 
at  0°  =  1  .4  X  10-60  mm.  Hg. 

SILVER. 

Sm  (0°—  1°)  up  to  400°  =  0.0555  +0.00000943t 

(Pionchon)  . 
over  400°  =  0.05758+0.0000044t+ 

0.000000006t2  (Pionchon). 
Q  (solid,  at  M.  P.  =  962°)        =  64.8  Cal. 
L.  H.  Fusion  =  24.35  Cal.  (Pionchon). 

Q  (liquid,  at  M.  P.)  =  89.15  Cal.  (Pionchon), 

S  (liquid,  962°  to  1020°)  =  0.0748  (Pionchon). 

Q  (liquid  at  B.  P.760  =  2040°)    =  170  Cal. 
L.  H.  Vaporization  (2040°)         =  490  Cal.  (calculated  from 

vapor  tension  data)  . 

Q  (vapor,  at  2040°)  =  660  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.046  (calculated). 


1  1 

Vapor  tension,  liquid  :     log  p.  =  --  Jp  --  f-  7.9  (from  experi- 

mental data). 
at  M.  P.  =  3.2  X10-2  mm.  Hg. 


19 
solid  :       log  p  =  -  ^         +  8.36. 

at  0°  C.  =  5.6  X10~37  mm.  Hg. 

CADMIUM. 

Sm  (0°—  1°)  =  0.0546+0.000012t  (Naccari). 

Q  (solid,  at  M.  P.  =  321°)         =  18.8  Cal. 

L.  H.  Fusion  =  13.0  Cal.  (Person). 


94  METALLURGICAL  CALCULATIONS. 

w 

Q  (liquid,  at  M.  P.)  =  31.8  (Person). 

S  (liquid)  =  0.0623  (calculated  same  as 

solid,  at  M.  P.). 

Q  (liquid,  at  B.  P.760  =  778°)     =  60.3  Cal. 
L.  H.  Vaporization  (778°)  =  251  Cal.  (calculated  from 

vapor  tension  curve) . 

Q  (vapor,  at  778°)  =  311  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed), 

per  kilo.  =  0.045  (calculated). 

£*~t   £  ~[ 

Vapor  tension,  liquid  :      log  p  = ™ — \-  8.74   (from    experi- 
mental data), 
at  M.  P.             =  2.6XlO-2mm. 

6470 
solid  :       log  p  =  -  ^-^  +  9.28. 

at  0°  C.  =  3.8  X 10-15  mm.  Hg. 

TIN. 

Sm  (0°— 1°)  =  0.0560+0.000044t  (Bede, 

combined  with  Regnault.). 
Q  (solid,  at  M.  P.  =  232°)         =  14.34  Cal. 
L.  H.  Fusion  =  13.82  Cal.  (Richards). 

Q  (liquid,  at  M.  P.)  =  26.16  Cal.  (Richards). 

Sm  (0°— t°)(t  =  232°  to  1000°)  =  0.06129-  0.00001047t+ 

0.00000001035t2  +  ^^ 

b 

(Pionchion) . 
S  (liquid)  at  t  =  232°  to  1000°  =  0. 06 129-0. 00002094t+ 

0.00000003 105t2  (Pionchon). 
Q  (liquid,  at  B.  P.760  =  2250°)  =  157  Cal. 
L.  H.  Vaporization  (2250°)         =  538  Cal.  (calculated  from 

Trouton's  rule,  K  =  25.1). 
Q  (vapor,  at  2250°)  =  695  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed), 

per  kilo  =  0.042  (calculated). 

13  930 

Vapor  tension,  liquid  :     log  p  = ^ 1-8.4  (Constants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  6.3XlO-20mm.  Hg. 

solid:       logp  =  -^|??  +  9.1. 
at  0°  C.  =  5.6  X  10~44  mm.  Hg. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        95 

ANTIMONY. 

Sm  (0°— 1°)  =  0.04864+0.0000084t 

(Naccari) . 

Q  (solid,  at  M.  P.  =  630°)         =  34  Cal. 
L.  H.  Fusion  .       =  40.3  Cal.  (Richards). 

Q  (liquid,  at  M.  P.  )  =  74.3  Cal.  (Richards). 

S  (liquid,  632°  to  830°)  =  0.0605  (Richards). 

Q  (liquid,  at  B.  P.76o  =  1500°)  =  127  Cal. 
L.  H.  Vaporization  (1500°)         =  124  Cal.  (calculated  from 

Trouton's  rule  for  Sbs,  K  = 

25.1). 

Q  (vapor,  at  1500°)  =  251  Cal. 

S  (vapor)  per  cubic  meter  =  0.405  (assumed,  for  vapor  = 

Sb3). 
per  kilo  =  0.025  (calculated  for  Sb8). 

9790 
Vapor  tension,  liquid  :      log  p  = ™ — \-  8.4  (constants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  3.55XlO-3mm.  Hg 

solid  :       log  p  =  -  ^0  +  9.58. 
at  0°  C.  =  6 . 5  X  10~31  mm.  Hg. 

IODINE. 

Sm  (9°— 98°)  -  =  0.05412  (Regnault). 

Q  (solid,  at  M.  P.  =  114°)         =  6.2  Cal. 
L.  H.  Fusion  =  11.7  Cal.  (Person). 

Q  (liquid,  at  M.  P.)  =  17.9  Cal. 

S  (liquid)  =  0.0676  (assumed). 

Q  (liquid,  at  B.  P.760  =  185°)     =  22.7  Cal. 

L.  H.  Vaporization  (185°)  =  23.95  Cal.  (Fabre  and  Silber- 

mann) . 

Q  (vapor,  at  185°)  =  46.65  Cal.  per  kg. 

S  (vapor)  per  kilo  (206°  to  377°)  =  0.034  (Strecker). 

per  cubic  meter  =  0.389  (calculated). 

percu.  meter  ( >  1200°)  =  0.225  (assumed  for  I  gas). 

per  kilo  =  0.039  (calculated). 

L.  H.  Change  of  State  (I2  =  21)  =  145  Cal.   per  kilo  (calculated 

from  vapor    tension  curve). 
=  1658  Cal.  per  cubic  meter  12. 


96  METALLURGICAL  CALCULATIONS. 

Q  (vapor,  at  1200°  =  I  vapor)    =  226  Cal.  per  kilo. 

2390 

Vapor  tension,  liquid  :      log  p  =  --  ™  —  [-8.10  (from  experi- 

mental data). 
at  M.  P.  =  89  mm.  Hg. 


solid  :       log  p  =  -p  +  10.43. 

atO°  =  2XlO-2mm.  Hg. 

i 

TELLURIUM. 

Sm  (0°—  455°)  =  0.0613  (Richards). 

Sm  (0°—  1°)  =  0.0495+0.000026t  (Richards). 

Q  (solid,  at  M.  P.  =  455°)          =  27.3  Cal. 

L.  H.  Fusion  =  19.0  Cal.  (Richards). 

Q  (liquid,  at  M.  P.)  =  46.3  Cal.  (Richards). 

S  (liquid)  =  0.0731  (calculated  same  as 

solid  at  M.  P.) 

Q  (liquid  at  B.  P.760  =  1390°)    =  115  Cal. 

L.  H.  Vaporization  (1390)°         =  166  Cal.  (calculated  by  Trou- 

ton's  rule  for  Te2*, 
K  =  25.1). 

Q  (vapor,  at  1390°)  =  281  Cal. 

S  (vapor)  per  cubic  meter          =  0.315  (assumed  for  Te2) 
per  kilo  =  0.028  (calc.  for  Te2). 

Vapor  tension,  liquid  :      log  p  =  --  ^—+8.4   (constants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  6.3X10~5  mm.  Hg. 

solid:       logp=  -i°|?0+9.84. 
at  0°  C.  =  8.1  X  10~32  mm.  Hg. 

CAESIUM. 

Sm  (0°—  26°)  =  0.0482  (Eckhardt  and  Graefe). 

Q  (solid,  at  M.  P.    =  29°)          =1.4  Cal. 

L.  H.  Fusion  =  4.8  Cal.  (calculated  by  2.1  T 

rule). 

Q  (liquid,  at  M.  P.)  =  6.2  Cal. 

L.  H.  Vaporization  (670°)  =  163  Cal.  (calculated  by  Trou- 

ton's rule;  K  =  23). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         97 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.038  (calculated). 

4760 
Vapor  tension,  liquid  :      log  p  =  --  ^-+7.92  (constants  from 

Trouton's  rule;  K  =  23). 
at  M.  P.  =  1.2XlO-8mm.  Hg. 

4QOO 
solid  :       log  p  =  — 


at  0°  C.  =  2.5  X  10-10  mm.  Hg. 

BARIUM. 

Sm  (0°—  100°)  =  0.05  (Mendeleeff). 

Q  (solid,  at  M.  P.  =  850°)         =  42.5  Cal. 
L.  H.  Fusion  =  17.0  Cal.  (calculated  by  2.1  T 

rule). 
Q  (liquid,  at  M.  P.)  =  59.5  Cal. 

CERIUM. 

Sm  (0°—  100°)  =  0.0448  (Hillebrand)  . 

Q  (solid,  at  M.  P.  =  623°)         =  27.9  Cal. 
L.  H.  Fusion  =  13.4  Cal.  (calculated  by  2.1  T 

rule). 

Q  (liquid,  at  M.  P.)  =  41.3  Cal. 

S  (liquid)  =  0.07  (assumed). 

TANTALUM. 

Sm.  (0°—  100°)  =  0.0354  (assumed,  from  at.  sp. 

heat  =  6.4). 

Sm  (0°—  1°)  =  0.0338+0.  000016t     (function 

of  t  assumed). 

Q  (solid,  at  M.  P.  =  2900°)        =  233  Cal. 

L.  H.  Fusion  =  37  Cal.  (calculated  by  2.1  T 

rule). 
Q  (liquid,  at  M.  P.)  =  270  Cal. 

TUNGSTEN. 

Sm  (0°—  1°)  (to  423°)  =  0.033  +0.00001  It    (Defacqz 

and  Geuchard). 

Q  (solid  at  M.  P.  =  3350°)        =  234  Cal. 

L.  H.  Fusion  =  40  Cal.  (calculated  by  2.1  T 

rule). 


98  METALLURGICAL  CALCULATIONS. 

Q  (liquid,  at  M.  P.)  =  274  Cal. 

S  (liquid)  =  0.10  (assumed,  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.76o  =  5164°)  =  455  Cal. 

L.  H.  Vaporization  (5164°)         =  743  Cal.   (B.  P.  according  to 

Langmiur)  . 

Q  (vapor,  at  5164°)  =  1198  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed). 
per  kilo  -  0.027  (calculated). 

30  000 

Vapor  tension,  liquid  :      log  p  =  --  ^  --  h8.4  (constants  from 

Trouton's  rule,  K  =  25.1). 
at  M.  P.  =  1.3  mm.  Hg. 

solid  :       log  p  =  - 


.. 
at  0°  C.  =  7.4  X  10-108  mm.  Hg. 

OSMIUM. 

Sm  (19°—  98°)  =  0.03113  (Regnault). 

Sm  (0°—  1°)  =  0.0305+0.000006t  (function  of 

t  assumed). 

Q  (solid,  at  M.  P.  =  2200°)         =  96  Cal. 

L.  H.  Fusion  =  36  Cal.  (calculated  by  second 

rule). 

Q  (liquid,  at  M.  P.)  =  132  Cal. 

S  (liquid)  =  0.057  (assumed  from  sp.   ht. 

solid  at  M.  P.). 

Q  (liquid,  B.  P.760  =  2600°)       =  155  Cal. 
L.  H.  Vaporization  (2600°)         =  379  Cal.  (calculated  from 

Trouton's  rule,  K  =  25.1). 
Q  (vapor,  at  2600°)  =  534  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed). 
per  kilo  =  0.026  (calculated). 

Vapor  tension,  liquid  :      log  p  =  --  ^  --  [-8.4  (constants  from 

Trouton's  rule;  K  =  25.1). 
at  M.  P.  =  97.7  mm.  Hg. 

solid:      logp=  -±M8_°+8.81. 

at  0°  C.  =  9.6  X10~54  mm.  Hg. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.         99 

IRIDIUM. 

Sm  (0°—  1°)  =  0.0317+0.000006t  (Violle). 

Q  (solid,  at  M.  P.  =  2360°)         =  108  Cal. 

L.  H.  Fusion  =  28  Cal.  (calculated  by  second 

rule). 

Q  (liquid,  at  M.  P.)  =  136  Cal. 

S  (liquid)  =  0.060  (assumed,  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.76o  =  2550°)  =  147  Cal. 
L.  H.  Vaporization  (2550°)         =  368  Cal.  (calculated  from 

Trouton's  rule;  K  =  25.1). 
Q  (vapor,  at  2550°)  =  515  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.020  (calculated). 

Vapor  tension,  liquid  :      log  p  =  --  7p  --  f-8.4  (constants  from 

Trouton's  rule;  K  =  25.1). 
at  M.  P.  =  302  mm.  Hg. 


solid  :       log  p  =  - 

at  0°  C.  =  5.6  X  10~53  mm.  Hg. 

PLATINUM. 

Sm  (0°—  1°)  =  0.0317+0.000006t  (Violle). 

Q  (solid,  at  M.  P.  =  1755°)       =  75.2  Cal.  (Violle). 
L.  H.  Fusion  =  27.2  Cal.  (Violle). 

Q  (liquid  at  M.  P.)  =  102.4  Cal.  (Violle). 

S  (liquid)  =  0.053  (assumed,  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.76o  =  2450°)  =  139  Cal. 
L.  H.  Vaporization  (2450°)         =  351  Cal.  (calculated  from 

Trouton's  rule;  K  =  25.1). 
Q  (vapor,  at  2450°)  =  490  Cal. 

S  (vapor)  per  cubic  meter  =  0.225  (assumed). 

per  kilo  =  0.026  (calculated). 

Vapor  tension,  liquid  :      log  p  =  --  7p  --  h8.4  (constants  from 

Trouton's  rule;  K  =  25.1). 
at  M.  P.  =  9.55  mm.  Hg. 

solid:       logp  =  -i^_°+9.0. 

at  0°  C.  =  3.0  X  10~51  mm.  Hg. 


100 


METALLURGICAL  CALCULATIONS. 


S  (0°— 600°) 

Sm  (0°— 1°)600°  to  1064° 

Q  (solid,  at  M.  P.  =  1064°) 

L.  H.  Fusion 

Q  (liquid,  at  M.  P.) 

S  (liquid) 

Q  (liquid,  at  B.  P.760  =  2530°) 
L.  H.  Vaporization  (2530°) 

Q  (vapor,  at  2530°) 
S  (vapor)  per  cubic  meter 
per  kilo 


GOLD. 
=  0.0316  constant  (Violle). 


Qfi 


=  0.0289+0.0000045t+ 

j 

(Violle). 

=  34.63  Cal.  (from  Violle's  equa- 

tion) . 

=  16.30  Cal. 

=  50.93  Cal.     (Roberts-Austen). 
=  0.0358  (assumed,  same  as  solid 
at  M.  P.). 
=  103.4  Cal. 
=  358  Cal.  (calculated  from 

Trouton's  rule;  K  =  25.1). 
=  461  Cal. 
=  0.225  (assumed). 
=  0.025  (calculated). 


Vapor  tension,  liquid  :      log  p  =  — 


15,470 


+8.4  (constants  from 


at  M.  P. 
solid  : 
at  0°  C. 


Trouton's  rule,  K  =  25.1). 
=  6.8XlO-4mm.  Hg. 


log  p  =  - 


16,175 


+8.93. 


=  4.8XlO-51mm.  Hg. 
MERCURY. 

=  0.0319  (Regnault). 

=  -4.1  Cal. 

=  2.84  Cal.  (Person). 

=  -1.30  Cal. 

=  0.0333  (Pettersson). 

=  0.03337 -0.0000027t+ 

0.00000000055t2  (Naccari). 
Q  (liquid,  at  B.  P.760  =  357°)     =  11.7  Cal. 


S  (solid)  at  -59° 
Q  (solid)  at  M.P.  -39° 
L.  H.  Fusion  (-39°) 
Q  (liquid,  at  M.  P.) 
Sm  (liquid,  -36°  to  0°) 
Sm  (0°— 1°)  up  to  250° 


L.  H.  Vaporization  (357°) 
(357°) 
(0°) 


L.H.V.  at  critical  temperature 
(1139°) 


=  67.8  Cal.  (Kurbatoff). 


=  66.8  Cal. 
=  77.14  Cal. 


0.0  Cal. 


(Calc.    thermo- 
dy  n  ami  call  y, 
from  vapor  ten- 
sion curve). 
1139°   is    calcu- 
lated critical 
temperature. 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.      101 

Q  (vapor,  at  357°)  =  79.5  Cal. 

(vapor,  at  1139°  =  critical  temp.)  =  100  Cal. 
S  (vapor)  per  cubic  meter.          =  0.225  (assumed), 
per  kilo  =  0.025  (calculated). 

Q  Q  Q  r\      f* 

Vapor  tension,  liquid  :      log  p  = ^r~  +  1.75  log.  T  — 

0.00221T+4.6544  (from  vapor 
tensions  at  357°,  260°  and  140°) . 
at  M.  P.  =  5.13X10-6mm.  Hg. 

solid  :       log  p  =  -?^+8.26  (8.26  assumed). 

THALLIUM. 

Sm  (17° -100°)  =  0.03355  (Regnault). 

Sm  (0°— 1°)  =  0.030+0.00002t. 

Q  (solid,  at  M.  P.  =  303°)         =  10.9  Cal.  (calculated  from  for- 
mula) . 

L.  H.  Fusion  =  7.2  Cal.  (Robertson). 

Q  (liquid,  at  M.  P.)  =  18.1  Cal. 

S  (liquid)  =  0.042  (assumed,  from  solid  at 

M.  P.). 

Q  (liquid,  at  B.  P.760  =  1700°)  =  77  Cal. 

L.  H.  Vaporization  (1700°)       =221  Cal.  (calculated by  Trouton's 

rule;  K  =  23). 

Q  (vapor,  at  1700°)  =  298  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed), 
per  kilo  =  0.025  (calculated). 

9  940 

Vapor  tension,  liquid  :     log.  p  =  —  ^p h  7.92  (constants  from 

Trouton's  rule;  K  =  23). 
at  M.  P.  =  4.7XlO-10mm.  Hg. 

r  A         1                  10,260  ,  0  _ 
solid:       logp  = ^ h-8.5. 

at  0°  C.  =  8.3  X 10-30  mm.  Hg. 

LEAD. 

Sm  (0°— 1°)  =  0.02925  +  0.000019t(Bede,  com- 

bined with  Regnault). 

Q  (solid,  at  M.  P.  =  327°)         =  11.6  Cal.  (Le  Verrier). 
L.  H.  Fusion  =  6.0  Cal.  (Richards). 

Q  (liquid,  at  M.  P.)  =  17.6  Cal.  (Person). 

S  (liquid,  335°  to  430°)  =  0.0402  (Person). 


102  METALLURGICAL  CALCULATIONS. 

Q  (liquid,  at  B.  P.76o  =  1580°)  =  68  Cal. 

L.  H.  Vaporization  =  209  Cal.  (calculated  from  vapor 

tension  curve). 

Q  (vapor,  at  1580°)  =  277  Cal. 

S  (vapor)  per  cubic  meter          =  0.225  (assumed), 
per  kilo  =  0.024  (calculated). 

Vapor  tension,  liquid:    log.  p  = ™ — h    8.0    calculated    from 

Greenwood's  data), 
at  M.  P.  =  1.5XlO-8mm.  Hg. 

9744 
solid:       logp  = ^-+8.41. 

at  0°  C.  =  4.9  X10-28  mm.  Hg. 

BISMUTH. 

Sm  (0°— 1°)  =  0.0285+0.00002t    (Bede,  com- 

bined with  Regnault). 
Q  (solid,  at  M.  P.  =  269°)         =  9.0  Cal. 
L.  H.  Fusion  =  12.0  Cal. 

Q  (liquid,  at  M.  P.)  =  21.0  Cal.  (Person). 

Sm  (liquid,  280°  to  360°)          =  O.Q363  (Person). 
Q  (liquid,  at  B.  P. 760  =  1435°)  =  63  Cal. 

L.  H.  Vaporization  (1435°)       =  208  Cal.  (calculated  from  vapor 

tension  curve). 

Q  (vapor,  at  1435°)  =  271  Cal. 

S  (vapor)  per  cubic  meter         =  0.225  (assumed), 
per  kilo  =  0.024  (calculated). 

9460 

Vapor  tension,  liquid  :      log  p  = ™ — [-7.45  (from  experimen- 
tal data). 
at  M.  P.            =  1 X 10-10  mm.  Hg. 

r,                           10,000   .      AK 
solid  :     log  p  = i=p h  8.45. 

at  0°  C.  =  6.6  X  10~29  mm.  Hg. 

RADIUM. 

Sm  0°  =  0.027  (from  Dulong  and  Petit 's 

Law). 

Sm  (0°-t°)  =  0.027 +0.000015t  (coefficient  of 

t  assumed). 

Q  (solid,  at  M.  P.  =  700°)       =  26.3  Cal. 

L.  H.  Fusion  =  9.0  Cal.   (calculated  by  2.1  T 

rule). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.      103 

Q  (liquid,  at  M.P.)  =  35.3  Cal. 

S  (liquid)  =  0.048  (assumed,  same  as  solid 

at  M.  P.). 

S  (vapor)  per  cubic  meter        =  0.225  (assumed). 
per  kilo  =  0.022  (calculated). 

THORIUM. 

Sm  (0°-100°)  =  0.0276  (Nilson). 

Sm  (0°-t°)  =  0.0270+0.000008t 

(coefficient  of  t  assumed). 
Q  (solid,  at  M.  P.  =  1690°)      =  68.5  Cal. 

L.  H.  Fusion  =  18.0  Cal.  (calculated  by  2.1  T 

rule). 

Q  (liquid,  at  M.  P.)  =  86.5  Cal. 

S  (liquid)  =  0.054  (assumed,  same  as  solid 

at  M.  P.). 
S  (vapor)  per  cubic  meter         =  0.225  (assumed) 

per  kilo  =  0.022  (calculated). 

URANIUM. 

Sm  (0°-98°)  =  0.028  (Blumcke). 

Sm  (0°-t°)  =  0.0275  +0.000007t  (coefficient 

of  t  assumed). 

Q  (solid,  at  M.  P.  =  2000°)      =  83  Cal. 

L.  H.  Fusion  =  20  Cal.    (calculated  by   2.1  T 

rule). 

Q  (liquid,  at  M.  P.)  =  103  Cal. 

S  (liquid)  =  0.055   (assumed,  same  as  solid 

at  M.  P.). 

Q  (liquid,  at  B.  P.760  =  2900°)  =  152  Cal. 

L.  H.  Vaporization  (2900°)       =   333     Cal.      (calculated     from 

Trouton'srule;  K  =  25.1). 
Q  (vapor,  at  2900°)  =  485  Cal. 

S  (vapor)  per  cubic  meter         =  0.225  (assumed). 
per  kilo  =  0.021  (calculated). 

Vapor  tension,  liquid  :    log  p  =  --  ^  --  h  8.4  (constants  from 

Trouton's  rule;  K  =  25.1). 
at  M.  P.  =  5.5  X  10~6  mm.  Hg. 


solid:       logp=  -- 

atO°  =  lX10-57mm.  Hg, 


104  METALLURGICAL  CALCULATIONS. 

in  several  of  the  preceding  instalments  we  have  given  the 
heats  of  formation  of  alloys  and  compounds  and  the  thermo- 
physics  of  the  elements.  Before  passing  to  the  thermophysics 
of  alloys  and  compounds  and  problems  involving  their  use, 
we  will  consider  a  few  simple  cases  of  the  application  of  data  so 
far  given.  Such  include  operations  in  which  metals  are  melted 
or  volatilized,  or  amalgams  retorted.  A  few  words  may  be  in 
order,  to  clear  the  ground,  regarding  what  is  to  be  regarded  as 
the  efficiency  of  a  furnace. 


Efficiency  of  Furnaces. 

Under  this  term  we  must  distinguish  a  generic  sense  and  a 
specific  sense,  the  first  referring  to  furnaces  in  which  the  object 
is  to  maintain  a  certain  temperature  for  a  certain  time  with 
the  minimum  consumption  of  fuel,  the  second,  in  which  the 
object  is  to  perform  a  certain  thermal  operation  with  the  small- 
est consumption  of  fuel.  In  the  first  case,  one  furnace  may  be 
compared  with  another,  and  thus  comparative  efficiencies  cal- 
culated; in  the  second  case  real  or  absolute  efficiencies  can  be 
also  calculated.  A  few  examples  will  illustrate  this  difference, 
which  is  an  essential  difference  as  far  as  making  calculations  is 
concerned. 

Cases  of  Specific  Efficiency:  Whenever  it  is  desired  to  melt 
a  metal  for  the  purpose  of  casting  it,  a  certain  definite  amount 
of  heat  must  be  imparted  to  the  metal,  and  the  ratio  between 
this  efficiently  utilized  heat  and  the  heating  power  of  the  fuel 
consumed,  is  the  efficiency  of  the  furnace.  If  the  furnace  is 
electric  the  theoretical  heat  value  of  the  electric  energy  used 
is  the  divisor.  If,  in  addition  to  the  heat  required  to  raise 
the  substances  to  the  desired  temperature,  there  is  also  heat 
absorbed  in  chemical  reactions,  this  amount  can  be  added  in 
as  usefully  applied  heat,  and  the  sum  of  this  and  the  heat  in 
the  final  products  be  regarded  as  the  total  efficiently  applied 
heat.  If  a  blast  furnace  takes  iron  ore  and  furnishes  us  melted 
pig  iron,  the  sum  of  the  heat  absorbed  in  the  chemical  decom- 
position of  the  iron  oxide  and  the  sensible  heat  in  the  melted 
pig  iron  is  the  efficiently  applied  heat,  because  it  is  the  neces- 
sary theoretical  minimum  required ;  all  other  items  are  more 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.      105 

or  less  susceptible  of  reduction,  but  these  are  necessary  items 
and,  therefore,  measure  the  net  efficiency.  If  my  dwelling  re- 
quires 200  cubic  feet  of  hot  air  per  minute  at  150°  F.  to  keep 
it  at  65°  F.,  while  the  outside  air  is  at  0°  F.,  the  ratio  of  the 
heat  required  to  warm  the  200  cubic  feet  of  air  from  0°  F.  to 
150°  F.,  to  the  calorific  power  of  the  fuel  used  per  minute, 
measures  the  specific  efficiency  of  the  "heater;"  the  question 
whether  this  amount  of  hot  air  keeps  the  temperature  of  the 
rooms  at  65°  F.  is  a  question  of  the  general  efficiency  of  the 
construction  of  the  house. 

Cases  of  Generic  Efficiency.  Such  are  those  in  which  prac- 
tically all  the  heat  generated  eventually  leaves  the  furnace  by 
radiation  or  conduction,  or  useless  heat  in  waste  gases;  this  is 
the  case  when  a  certain  temperature  has  to  be  continuously 
maintained  for  a  given  time,  and  where  the  time  element  is  the 
controlling  one,  and  not  any  definite  amount  of  thermal  work 
is  to  be  done.  Examples  are  numerous:  An  annealing  furnace, 
where  steel  castings,  let  us  say,  are  to  be  kept  at  a  red  heat 
for  two  days,  or  a  brick  kiln,  where  several  days  slow  burning 
are  required,  or  a  puddling  furnace,  where  the  melted  iron 
must  be  held  one  to  two  hours  to  oxidize  its  impurities.  In 
all  these  cases  we  may  say  that  one  furnace  keeps  its  con- 
tents at  the  right  heat  for  the  right  time  with  so  much  fuel, 
another  does  the  same  work  with.  10  or  25  per  cent,  less  fuel, 
and  is,  therefore,  10  or  25  per  cent,  more  efficient;  but  we  can- 
hot,  in  the  nature  of  the  case,  speak  of  the  absolute  or  specific 
efficiency  of  the  furnace,  because  there  is  no  definite  term, 
expressible  in  calories,  to  compare  with  the  thermal  power  of 
the  fuel. 

In  many  cases  the  two  efficiencies  are  mixed  in  the  same 
process  or  operation,  and  then  the  calculation  of  absolute  or 
specific  efficiency  can  be  made  for  that  portion  of  the  operation 
wherein  a  certain  definite  amount  of  thermal  work  is  done. 
Thus,  in  an  annealing  kiln,  50  tons  of  castings  may  be  brought 
up  to  annealing  heat  in  24  hours,  starting  cold,  and  the  heat 
absorbed  by  the  castings  compared  with  the  calorific  power  of 
the  coal  burnt  during  this  period,  is  a  measure  of  the  real  effi- 
ciency of  this  part  of  the  operation.  During  the  rest  of  the 
operation,  while  the  castings  are  simply  kept  at  annealing  heat, 


106  METALLURGICAL  CALCULATIONS* 

there  can  be  no  calculation  of  the  absolute  or  specific  efficiency 
of  the  furnace,  because  one  of  the  terms  necessary  for  the 
comparison  has  disappeared,  in  that  part  of  the  process  we 
can  only  speak  of  relative  efficiency  compared  to  some  other 
furnace  doing  a  similar  operation. 

It  goes,  almost  without  saying,  that  we  can,  of  course,  apply 
the  conception  of  efficiency  in  its  relative  or  general  sense  to 
the  whole  operation  or  to  any  part  of  it 

Problem  6. 

The  Rockwell  Engineering  Co.  state  in  their  current  adver- 
tisements that  their  regenerative  oil-burning  furnace  melts  100 
pounds  of  copper  with  the  consumption  of  less  than  1.5  gallons 
of  oil.  Assume  that  1.5  gallons  of  oil  is  used,  and  that  the 
copper  is  heated  from  25°  C.  to  melted  metal  100°  C.  above  its 
melting  point. 

Required:  The  "efficiency"  of  the  furnace;  i.e.,  its  specific 
efficiency  as  calculated  from  the  net  heat  utilized. 

Solution:  One  gallon  of  fuel  oil  averages  in  weight  7.5  pounds, 
and  its  calorific  power  11,000  Calories  per  kilogram,  or  11,000 
pound  Calories  per  pound.  The  calorific  power  of  the  fuel 
used  in  melting  100  pounds  of  copper  is  therefore: 

Heat  generated  11,000X7.5X1.5  =  82,500X1.5  <=  123,750 
pound.  Calories. 

The  heat  imparted  to  the  copper  is  as  follows,  taking  the 
data  from  Article  V.  of  these  calculations:  (p.  68.) 

Heat  in  1  Ib.  melted  copper  at  melting  point      =  162  Ib.  Cal. 

Heat  in  1  Ib.  solid  copper  at  25°  C.  = 2 

Heat  required  to  just  melt  the  copper                 =  160       " 
Heat    to    superheat    liquid    copper    100°    C. 

0.133X100=  13       " 

Total  heat  expended  on  each  pound  of  copper  =  173 

Heat  usefully  applied  per  100  pounds                 =  17,300 

1 7  onn 
Net  efficiency  of  furnace  =  ^3750°  °'14  =  14% 

It  is  proper  to  remark  that  although  this  efficiency  appears 
low,  yet  it  is  considerably  greater  than  is  attained  in  simple 
melting  holes  or  wind  furnaces,  and  yet  the  calculations  show 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.        107 

what  a  large  margin  for  improvement  and  greater  efficiency 
exists  in  even  some  of  the  best  and  relatively  most  efficient 
metallurgical  furnaces. 

Problem  7. 

In  the  distillation  of  silver  amalgam  in  iron  retorts,  1000 
kilos,  of  amalgam,  containing  200  kilos,  of  silver,  is  retorted 
with  the  consumption  of  550  kilos,  of  wood,  the  mercury  vapor 
passes  off  at  an  average  temperature  of  450°  C.,  and  the  silver 
is  raised  towards  the  end  of  the  operation  to  800°  C.,  in  order 
to  expel  the  last  of  the  mercury.  Assume  the  calorific  power 
of  the  wood  3000  Calories. 

Required:     The  net  efficiency  of  the  furnace. 

Solution:  The  heating  power  of  the  wood  is  550X3000  = 
1,650,000  Calories. 

The  heat  utilized  is  that  absorbed  in  separating  the  silver 
from  the  mercury  plus  the  sensible  heat  in  the  mercury  vapor 
at  450°,  plus  the  sensible  heat  in  the  silver  at  800°.  These  are 
calculated  as  follows: 

Heat  to  decompose  amalgam  =  2470  Calories  per  108  kilos, 
of  silver  =  200  X  (2470  - 108)  =  200X22.9  =  4,580  Calories. 

Heat  in  silver  at  800°,  using  Pionchon's  formula,  is  800 
[0.05758  +  0.0000044  (800) +0.000000006  (800)2]X200  =  10,390 
Calories. 

Heat  in  800  kilos,  of  mercury  vapor  at  450°  is 

(a)  heat  to  boiling  point  (Naccari) 
357  [0.03337—0.00000275  (357)  + 

0.0000000667  (357)2]  X 800  .  =     11 ,677  Cal. 

(b)  heat  to  vaporize  72.5X800  =     58,000     " 

(c)  heat  in  vapor  at  450°  = 

0.025  X  (450— 357)  X  800  =       1,880     " 


Total  =  71,557  " 
Heat  usefully  applied : 

In  decomposing  amalgam  =  4,580  " 

In  mercury  vapor,  as  sensible  heat  =  71 ,557  " 

In  silver,  as  sensible  heat  =  10,390  " 


Total  =    86,527    « 

86  527 

Efficiency  of  furnace  =  1      '  nnr.  =  0.052  =  5.2%. 

1,DOU,UUU 


108  METALLURGICAL  CALCULATIONS. 

Problem  8. 

In  a  zinc  works,  impure  zinc  is  refined  by  redistillation  in 
fire-clay  retorts,  a  bank  of  retorts  distilled  970  kilos,  of  zinc 
with  the  expenditure  of  912  kilos,  of  small  anthracite  coal. 
Assume  that  the  zinc  vapors  pass  out  of  the  muffles  at  the 
boiling  point  (930°). 

Required:     (1)  The  net  efficiency  of  the  furnace. 

(2)  The  electrical  power  which  would  be  required,  in  horse- 
power-hours, to  do  the  same  work,  assuming  the  heating  effi- 
ciency of  the  electric  furnace  is  75  per  cent. 

Solution:  (1)  The  small  anthracite  may  be  assumed  to  have 
a  calorific  power  of  7850  Calories;  therefore,  the  total  heat 
which  should  be  developed  is  7850X912  =  7,159,200  Calories. 
The  heat  in  1  kilo,  of  zinc  in  the  state  of  vapor  at  its  boiling 
point  can  be  calculated  from  the  thermophysical  data  supplied 
for  zinc  as: 

(a)  In  solid  zinc  to  melting  point  (420°)  ...........    45.20  Cal. 

(6)  Latent  heat  of  fusion  ........................   22.61     " 

(c)  Heat  in  melted  zinc  to  boiling  point  (930°)  ____   65.05     " 

(d)  Latent  heat  of  vaporization  ...................  425.00    " 

Total        557.86    " 
Heat  required  for  970  kilos.  =  541,124    " 

541  124 
Efficiency  of  furnace  =  2QQ  =  0.075  =  7.5%. 


(2)  One  electric  horse-power-hour  =    644  .  0  Cal. 

Efficiently  applied  heat  =  644  X.  75  =    483.0    " 

489  995 
Electric  horse-power-hours  required  =  —  r~-  =  1013  E.H.P  hours 

4oo 

One  metric  ton  of  zinc  requires  =  1044  E.H.P.  hours. 


Cost  of  power,  at  $20.00  per  E.  H.  P.  year  =       TQQ  X  1044  = 

0.00228X1044  =  $2.38. 

0  912 

This  cost  of  electric  power  would  replace  the  use  of     '     ^ 

metric  tons  of  small  anthracite,  equal  to  a  cost  of  $2.53  for 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.       109 

electric  power  sufficient  to  replace  a  metric  ton  of  coal  for  this 
purpose. 

Many  other  examples  could  be  given  of  the  technical  use  of 
the  thermophysical  data  concerning  the  elements,  but  the 
problems  given  illustrate  the  methods  of  calculation. 

When  one  is  acquainted  with  some  of  the  ordinary  metal- 
lurgical operations,  such  as  melting  and  distilling  the  metals, 
it  is  surprising  to  notice  how  little  is  known  or  thought  of  the 
efficiency  or  lack  of  efficiency  of  the  furnaces  used.  One  man 
melts  100  pounds  of  metal  by  the  use  of  150  pounds  of  coal,  he 
builds  a  new  furnace  and  does  it  more  cheaply  by  using  100 
pounds  of  coke,  which  is  certainly  relatively  more  efficient; 
but  it  is  seldom  that  the  operator  knows  that  in  one  case  he 
is  getting  probably  only  7  per  cent,  efficiency  from  his  fuel 
and  in  the  other  case  only  10  per  cent.  It  is  the  knowledge 
of  these  absolute  efficiencies  which  tells  the  practical  man  just 
what  he  is  accomplishing,  and  shows  him  how  much  room 
there  still  remains  for  improvement. 


THERMOPHYSICS  OF  ALLOYS. 

There  does  not  exist,  in  technical  literature,  much  data  of 
this  nature  concerning  alloys.  There  is  here  a  wide  and  inter- 
esting field  for  metallurgical  research,  whose  cultivation  would 
yield  results  both  of  high  practical  and  high  theoretical  inter- 
est, and  yet  it  is  comparatively  untouched.  What  is  wanted 
is  complete  data  concerning  the  specific  heat  of  solid  and  liquid 
alloy,  and  latent  heat  of  fusion.  These,  combined  with  the 
determination  of  the  heat  evolved  in  the  alloying,  would  fur- 
nish a  sound  basis  for  a  practical  theory  of  alloys,  besides 
enabling  workers  with  these  alloys  to  control  the  efficiency  of 
their  furnaces  and,  in  general,  to  know  with  scientific  exactness 
what  they  are  accomplishing. 

ALLOYS  OF  TIN  AND  LEAD. 

Per  Cent,  of  Tin.                                 Sm.  Latent  Heat  of  Fusion. 

4.8  (Pb16Sn)  5.5  (Mazotto) 

10.2  (Pb5Sn)  8.0  at  307°  (Spring) 

12.5  (Pb4Sn)  8.3  at  292°  (Spring) 


110 


METALLURGICAL  CALCULATIONS. 
Sm. 


Per  Cent,  of  Tin.                                  Sm.  Latent  Heat  of  Fusion. 

16.0  (Pb3Sn) 9.1  at  289°  (Spring) 

22.2  (Pb2Sn)  9.5  at  270°  (Spring) 

7.9  (Mazotto) 

36.3  (PbSn)        0.04073  (12°*-99°)         11.6  at  241°  (Spring) 

(Regnault) 

9.4  (Mazotto) 

50.0  =  total  heat  to  0°  in    1   kilo. 

melted  metal     18.0  from  202°  (Ledebur) 

53.3  (PbSn2)       0.04507  (10°— 99°)         10.5  at  197°  (Mazotto) 

(Regnault) 

63.1  (PbSn3)  15.5  at  179°  (Spring) 

69.5  (PbSn4)  17.0  at  188°  (Spring) 

83.0  =  total  heat  to  0°  in  1  kilo. 

melted  metal     21.5  from  205°  (Ledebur) 

90.1  (PbSn16)  12.9  (Mazotto) 

ALLOYS  OF  TIN  AND  BISMUTH. 

Per  Cent,  of  Tin.                        Sm.  Latent  Heat  of  Fusion. 

6.7  (Bi8Sn)  11.4  Cal.  (Mazotto) 

22.1  (Bi2Sn)  . .  . 11.2  Cal.  (Mazotto) 

36.2  (BiSn)         0.0400  (20°— 99°)  11.6  Cal.  (Mazotto) 

(Regnault) 
53.1  (BiSn2)        solid,  0.0450  11.6  (Cal.  Mazotto) 

(Regnault) 
liquid,  0.0454 
(146°— 275°)  Person 

69.1  (BiSn4)  11.1    Cal.   at    140° 

(Mazotto) 

82.7  (BiSn8)  12.6  Cal.  (Mazotto) 

90.1  (BiSn16)  12.8  Cal.  (Mazotto) 

ALLOYS  OF  TIN  AND  ZINC. 
Per  Cent,  of  Tin.  Sm.  Latent  Heat  of  Fusion. 

78.4  (ZnSn2)  . . . ..         23.5  Cal.  (Mazotto) 

92.7  (ZnSn7)  .- 16.2   Cal.   at    197° 

(Mazotto) 

95.6  (ZnSn12)  16.3  Cal.  (Mazotto) 

97.3  (ZnSn20)  15.1  Cal.  (Mazotto) 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.       Ill 

ALLOYS  OF  TIN  AND  COPPER. 

Bell  metal  (20  per  cent,  tin)  Sm  (14°— 98°)  =  0.0862  (Reg- 
nault). 

Bronze  (15  per  cent.  tin). 

Total  heat  in  melted  metal  (to  0°)  =  130  Cal.  (Ledebur). 

If  strongly  superheated  =  143.5  Cal.  (Ledebur). 

ALLOYS  OF  TIN  AND  ANTIMONY. 

Britannia  metal  (90  per  cent,  tin  )  requires  to  melt  it,  starting 
cold,  28.0  Calories  per  kilogram  (melting  point  236°);  with  82 
per  cent,  of  tin,  25.7  Calories;  melting  point  205°  (Ledebur). 

ALLOYS  OF  TIN,  BISMUTH  AND  ANTIMONY. 

BiSn2Sb   (bismuth  34.3,  tin  41.9,  antimony  23.8  per  cent.). 
Sm  (15°— 100°)  =  0.0462  (Regnault). 

ALLOY  OF  TIN,  BISMUTH,  ANTIMONY  AND  ZINC. 

BiSn2SbZn2  (bismuth  29.8,  tin  34.0,  antimony  17.3,  zinc  18.9 
per  cent.). 

Sm  (15°— 100°)  =  0.0566  (Regnault). 

ALLOYS  OF  LEAD  AND  BISMUTH. 
Per  Cent,  of  Lead.  Sm.  Latent  Heat  of  Fusion. 

11.1  (PbBi8)  10.2  (Mazotto) 

33.2  (PbBi2)  6.4  (Mazotto) 

39.9  (Pb2Bi3)      solid,  0.03165 

(16°— 99°)  Person, 
liquid,  0.03500 
(144°— 358°)     Person 

42.7  (Pb3Bi4)  4.7  at  127°  (Mazotto) 

49.9  (PbBi)  4.0  (Mazotto) 

66.6  (Pb2Bi) 3.6  (Mazotto) 

88.8  (Pb'Bi)  4.9  (Mazotto) 

ALLOYS  OF  LEAD  AND  ANTIMONY. 

With  63.0  per  cent  of  lead,  Sm  (10°— 98°)  =  0.0388  (Regnault) 
With  82.0  per  cent,  of  lead,  heat  in  1  kilo,  melted  metal  = 

15.6  Calories    (Ledebur). 
With  90.0  per  cent,  of  lead,  total  heat  in  1  kilo,  melted  metal 

=  13.8  Calories  (Ledebur). 


112  METALLURGICAL  CALCULATIONS. 

ALLOYS  OF  LEAD,  TIN  AND  BISMUTH. 

D'Arcet's  Alloy,  containing  32.5  lead,  18.5  tin,  48.7  bismuth 
Sm  solid  (5°— 65°)  =  0.0372  (Mazotto) 

Sm  solid  (12°— 50°)  =  0.049  (Person). 

Sm  solid  (14°— 80°)  =  0.060  (Person). 

Sm  liquid  (107°— 136°)         =  0.047  (Person). 
Sm  liquid  (120°— 150°)         =  0.0399  (Mazotto). 
Sm  liquid  (136°— 300°)         =  0.0360  (Person). 
Latent  heat  of  fusion  =  5.96  Cal.  at  96°  (Person). 

=  5.77  Cal.  at  99°  (Mazotto). 

Rose's  Alloy,  containing  24.0  lead,  27.3  tin,  48.7  bismuth: 
Sm  solid  (5°— 65°)  =  0.375  (Mazotto). 

Sm  fluid  (119°— 338°)  =  0.0422  (Person). 

Latent  heat  of  fusion  =  6.85  Cal.  at  99°  (Mazotto). 

Fusible  Alloy,  containing  31.8  lead,  36.2  tin,  32.0  bismuth: 
Sm  solid  (18°— 52°)  =  0.0423  (Person). 

Sm  solid  (11°— 98°)  =  0.0448  (Regnault). 

Sm  fluid  (143°— 330°)  =  0.0460  (Person). 

Latent  heat  of  fusion  =  7.63  Cal.  at  145°  (Person). 

Wood's  Alloy,  containing  25.8  lead,  14.7  tin,  52.4  bismuth: 
7  cadmium: 

Sm  solid  (5°— 50°)  =  0.0352  (Mazotto). 

Sm  fluid  (100°— 150°)  =  0.0426  (Mazotto). 

Latent  heat  of  fusion  =  7.78  Cal.  at  75°  (Mazotto). 

Lipowitz's  Alloy,  containing  25.0  lead,  14.2  tin,  50.7  bismuth, 
10.1  cadmium: 

Sm  solid  (5°— 50°)  =  0.0345  (Mazotto). 

Sm  fluid  (100°— 150°)  =  0.0426  (Mazotto). 

Latent  heat  of  fusion  =  8.40  Cal.  at  75°  (Mazotto). 

ALLOYS  OF  COPPER  AND  ZINC. 

Red  Brass  S  at    0°    =  0.0899  (Lorenz) 

S  at  50°    =  0.0924  (Lorenz) 

(Copper  85%)        S  at  75°    =  0.0940  (Lorenz) 

Yellow  Brass          Sat    0°    =  0.0883  (Lorenz) 

at  50°    =  0.0922  (Lorenz) 

(Copper  65%)  at  175°=  0.0927  (Lorenz) 

Heat  in  1  kilo,  of  melted,  somewhat  superheated,  brass  =* 
130  Calories  (Ledebur). 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.      113 

ALLOYS  OF  COPPER,  ZINC  AND  NICKEL. 

German  Silver: 

(74Cu.  20Zn.  6Ni)  Sm(0— t)  =  0.0941  +0.0000053t 

(Tomlinson) 

ALLOYS  OF  COPPER  AND  ALUMINIUM. 
Copper  88.7%  Sm  (20°— 100°)  =  0.10432  (Luginin) 

ALLOYS  OF  SILVER  AND  PLATINUM. 
Silver  66.7%  Sm  (0— t)  =  0. 04726  +  0.0000 138t  (Tomlinson). 

ALLOYS  OF  MERCURY  AND  TIN. 

HgSn  (37.1%  Sn)  Sm  (—30°— 15°)  =  0.04083  (Schiiz) 
(—25°— 15°)  =  0.04218  (Schiiz) 
(  22°— 99°)  =  0.07294  (Regnault) 

HgSn  (54.1%  Sn)  Sm   (     25°— 99°)  =  0.06591  (Regnault) 
HgSn5(74.7%  Sn)  Sm   (—16°— 15°)  =  0.05039  (Schiiz) 

ALLOYS  OF  MERCURY  AND  LEAD. 

Pb  Hg  (50.9%  Pb)  Sm  (—69°— 20°)  =  0.03458  (Schuz) 

Sm  (     23°— 99°)  =  0.03827  (Regnault) 
Pb2Hg  (67.4%  Pb)  Sm  (—72°— 20°)  =  0.03348  (Schiiz) 

ALLOYS  OF  CADMIUM  AND  TIN. 

CdSn2  (67.8%  Sn)  Sm  (—77°— 20°)  =  0.05537  (Schuz) 
whence  we  have  Sm  (    0  —    t)  =  0.0557  + 

0.00000366t  (Schuz) 

ALLOYS  OF  IRON  AND  CARBON. 

Soft  Steel  (0.15%  carbon)  Sm  (20°— 98°)  =  0.1165  (Regnault). 

Hard  Steel  (1.00%  carbon)  Sm  (20°— 98°)  =  0.1175  (Regnault). 

Total  heat  in   1   kilo  melted  steel  at   1350°  =  300  Calories 
(Ledebur). 

Cast  Iron  (4.0%  carbon)  Sm  (0—1200°)  =  0.175. 
Sm  (0— 1°)  =  0.12-f  0.000046t 

Total  heat  in  1  kilo,  melted  at  1200°  =  245  Calories  (Ledebur). 

Total  heat  in  1  kilo,  coming  from  blast  furnace  =  250  to  325 
Calories  (Akermann). 


114  METALLURGICAL  CALCULATIONS. 

Problem  9. 

A  steel-melting  crucible  contains  110  pounds  of  steel,  which 
is  melted  in  a  wind  furnace  with  the  use  of  150  pounds  of  coke. 
Assume  the  coke  to  be  90  per  cent,  fixed  carbon,  and  the  steel 
to  be  superheated  100°  C.  above  its  melting  point. 

Req^lired:     The  net  efficiency  of  the  furnace. 

Solution:  The  calorific  power  of  the  coke  may  be  assumed 
as  90  per  cent,  that  of  pure  carbon,  and  therefore: 

=  1  50  X  (8100X0.90)  =  150X7290   =  1,093,500  Ib.  Cal. 
Heat  in  steel  at  melting  point: 

110X300  (Ledebur)  ==  33,000 
Heat  to  superheat  100°: 

110X100X0.15  (assumed)    =     1,650 

Total     34,650  Ib.  Cal. 


Efficiency  of  furnace  -  yj  -  0.032  =  3.2% 


Problem  10. 

A  Siemen's  regenerative  furnace  holds  eighteen  steel  cruci- 
bles, each  containing  100  pounds  of  steeL  Assume  that  the 
efficiency  of  utilization  of  the  heat  for  melting  the  steel  is  5 
per  cent.,  and  that  the  furnace  is  fed  by  natural  gas,  having  a 
calorific  power  of  512-pound  Calories  per  cubic  foot. 

Required:  The  number  of  cubic  feet  of  natural  gas  required 
per  furnace  heat  of  eighteen  crucibles  =  1800  pounds  of  cast 
steel. 

Solution:  Heat  in  steel  =  1800x315  =  567,000  Ib.  Cal. 


Heating  power  of  gas  required  =  ~-  =11,340,000  Ib,  Cal. 
Cubic  feet  of  gas  required  =  -  —  —  ^  --  =  22,150  cubic  feet. 

0  1  Ju 

Gas  required  per  ton  of  steel,  2000  Ibs.  =  24,610  cubic  feet. 
Cost  of  gas,  at  $0.08  per  1000  cubic  feet  =  $1.97. 

Problem  11. 

In  a  malleable-casting  foundry  the  pig  iron  is  melted    in   a 
reverbatory  air  furnace,  3000  kilos,  being  melted   in   two  hours 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.      115 

by  the  combustion  of  1200  kilos,  of  bituminous  coal,  having  a 
calorific  power  of  8500  Calories. 

Required:    The  melting  efficiency  of  the  furnace. 

Solution'.    Calorific  power  of  coal  used: 

1200X8500  =  10,200,000  Calories. 
Heat  in  melted  iron  at  foundry  heat: 

3000X250  (Ledebur)  =  750,000  Calories. 


Efficiency  of  furnace  =  =  0.0735  =  7.35% 


Problem  12. 

In  an  iron  foundry  cupola  14  metric  tons  of  pig  iron  are 
melted  in  one  hour,  using  1.5  tons  of  coke  (90  per  cent,  car- 
bon). The  gases  passing  away  contain  by  volume  CO  13  per 
cent.,  CO2  13  per  cent.,  nitrogen  74  per  cent.,  and  leave  the  cupola 
at  500°  C.  The  body  of  the  cupola  is  1.5  meters  in  diameter 
outside  and  4  meters  high. 

Required: 

(1)  The  net  melting  efficiency  of  the  cupola. 

(2)  The  proportion  of  the  calorific  power  of  the  coke  lost. 

(a)  By  the  sensible  heat  .of  the  hot  gases  escaping. 

(b)  By  the  imperfect  combustion  of  the  coke. 

(c)  By  radiation  from  bottom  and  walls  of  the  cupola. 

(3)  The  amount  of  heat  in  Calories  radiated,  on  an  average, 
from  each  square  meter  of  outside  surface  per  minute. 

Solution  : 

(1)  Calorific  power  of  the  coke; 

1500X0.90X8100.  =  10,935,000  Calories. 
Heat  in  melted  iron: 

14,000X250  (Ledebur)  =  3,500,000  Calories. 


Efficiency  of  melting  =  -  0.32  =  32% 

(2  a)   Weight  of  carbon  escaping  =  1500X0.90  =  1350  kilos. 


116  METALLURGICAL  CALCULATIONS. 


Volume  of  CO  and  CO2  escaping  «=  —  ^  -  2500  m* 
(Because  1  m3  of  either  gas  carries  0.54  kilos.  C.) 
Volume  of  escaping  gas  =  n  13  +  0  13 

Volume  of  nitrogen  (by  difference)  =  7115m8 
Sensible  heat  of  nitrogen  and  CO 

(71  15  +  1250)  X  [0.303  (500)  -f- 

0.000027  (500)2]  -  1,323,760  Cal. 
Sensible  heat  of  CO2 

1250  X  [0.37  (500)  +  0.00022  (500)2]       -     300.000    " 
Total  sensible  heat  in  gases  «  1,623,760    " 

Proportion  of  calorific  power  of  the  fuel  thus  lost: 

1'623'760   --01485=  1485<? 
-  10,935,000  '  L4'85%- 

(2,  b)  Volume  of  CO  escaping  «          1250  ma 

Calorific  power  of  this  gas  =•  1250X 

3062  -  3,827,500  Cal. 
Proportion  of  calorific  power  of  the  fuel  thus  lost: 

3,827,500 
3  10,935,000 

(2,  c)    Heat  in  pig  iron  -    3,500,000  Cal. 

Heat  in  waste  gases  •=    1,623,760  " 

Lost  by  imperfect  combustion  «=    3,827,500  " 

Accounted  for  8,951,260  " 

Calorific  power  of  the  coke  •  10,935,000  " 

Difference,  loss  by  radiation  =     1,983,740  " 
Proportion  of  calorific  power  of  the  fuel  thus  lost: 


(3)  Area  of  bottom  of  cupola  -  (1.5)2X0.7854  -  1.77m2 
Area  of  sides  of  cupola  =  1.5X3,14X4  -  18.85m2 
Total  radiating  area  —  20.62  m2 

Heat  radiated  per  sq.  m.  per  hour  «    *  2Q  f        -  96,200  CaL 


Heat  radiated  per  sq.  m.  per  min.  «•  —  TTT—      -«    1,603 


THERMOCHEMISTRY  OF  HIGH  TEMPERATURES.       117 

In  general,  we  may  state  that  the  net  efficiency  in  melting, 
etc.,  of  metals  is  very  various,  from,  say,  2  or  3  per  cent,  in  a 
crucible  steel-melting  wind  furnace  to  about  10  or  15  per  cent, 
in  reverberatory  furnaces,  20  to  30.  per  cent  in  regenerative 
open -hearth  furnaces,  30  to  50  per  cent,  in  shaft  furnaces,  where 
material  to  be  heated  and  fuel  burned  are  in  direct  contact  with 
each  other,  50  to  75  per  cent,  in  steam  boilers  and  hot-blast 
stoves,  and  60  to  85  per  cent,  in  large  electrical  furnaces. 


CHAPTER  V. 
THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS. 

The  metallic  oxides  are  the  most  important  compounds 
occurring  in  metallurgy,  together  with  the  oxides  of  hydrogen, 
carbon  and  of  sulphur,  arsenic  and  antimony,  which  are  formed 
during  combustion  and  in  roasting  operations.  We  do  not 
include  as  oxides  those  combinations  of  two  or  more  oxides 
in  chemical  proportions,  such  as  of  silica  with  metallic  oxides, 
which  form  oxygen  salts  or 'ternary  oxygen  compounds.  They 
will  be  separately  discussed.  In  the  following  lists  Sm,  as 
before,  means  the  mean  specific  heat  per  kilogram,  in  large 
Calories  (or  per  pound  in  Ib.  Cal.)  in  the  range  of  temperature 
given;  S  means  actual  specific  heat  at  the  temperature  given, 
and  Q  is  the  quantity  of  heat  absorbed  in  fusion  or  volatiliza- 
tion or  in  passing  through  any  designated  range  of  tempera- 
ture. Temperatures  are  given  only  in  centigrade  degrees,  ex- 
cepting that  /  signifies  Fahrenheit  temperature;  volumes  of 
gases  are  understood  as  being  at  0°  C.  and  760  m.m.  pressure, 
unless  distinctly  stated  otherwise, 

OXIDES. 
Hydrogen  Oxide,  H20: 

Ice     Sm  (— 78°— 0°)  =  0.463  (Regnault). 

(— 30°— 0°)  =  0.505  (Person). 

Water— Sm  (0°— 100°)  =  l  +  0.00015t  (Pfaundler). 

S      (at  15°)  =  1.0045  (Pfaundler). 

S     (at  0°)  =  1.0000  (assumed). 

In  all  metallurgical  calculations  it  will  be  sufficiently  exact 
to  assume  the  specific  heat  of  water  at  ordinary  temperatures 
as  unity.  In  heating  to  the  boiling  point,  from  zero,  101.5 
Calories  are  absorbed. 

118 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.          119 

Gas  Sm  (0°  to  t°) 

1  m3  up  to  2000°  =  0.34  +  0.00015t  (Mallard  and  Le 

Chatelier). 

1  m3  2000°— 4000°  =  0.48  +  0..00017t  (Vielle). 

1  ft3  up  to  2000°  =  0.021  +  0.000009t— Ib.  Cal. 

1  ft3  2000°— 4000°  =  0.030  +  0.00001  It— Ib.  Cal. 

1  kilo,  up  to  2000°  =  0.42  +  0.0001S5t— Cal. 

1  kilo.  2000°— 4000°-          =  0.59  +  0.00021t  Cal. 

1  Ib.  up  to  2000°  =  0.42  +  0.0001S5t— Ib.  Cal. 

1  Ib.  2000°— 4000°  =  0.59  +  0.00021t— Ib.  Cal. 

1  ft3  up  to  3600°  F.  =  0.021  +  0.000005*  B.  T.  U. 

1  ft3  3600°— 7200°  F.         =  0.030  +  0.000006*  B.  T.  U. 

1  Ib.  up  to  3600°  F.  =  0.42  +  0.000103*  B.  T.  U. 

1  Ib.  3600°— 7200°  F.  =  0.59  +  0.00012*  B.  T.  U. 
Above  figures  are  for  water  vapor  as  it  exists  in  almost  all 
metallurgical  problems,  as  true  gas  far  removed  from  its  max- 
imum tension  at  any  given  temperature.  When  the  water 
vapor  exists  as  saturated  steam  under  its  maximum  tension, 
as  in  a  steam  boiler,  it  is  recommended  to  consult  tables  giving 
the  total  heat  in  steam  at  any  temperature  condensing  to  liquid 
water  at  0°,  or  to  use  Regnault's  formula  for  the  same: 

Q  =  606.5 +  0.305t 

which  expresses  the  amount  of  heat  required  to  convert  1  kilo- 
gram of  water,  liquid  at  0°,  into  steam  at  its  maximum  pressure 
at  the  temperature  t.  A  little  reflection  will  show  that  the 
above  formula  amounts  to  taking  the  specific  heat  of  a  kilo- 
gram of  saturated  steam,  under  a  constantly  increasing-pressure, 
as  Q  =  0.305t 

or  under  increasing  pressure  Sm  =  0.305 

or  per  cubic  meter  (0.81  kilos.)  Sm  =  0.247 

These  figures  apply  to  boiler  practice  only.  In  cases  where  water 
is  evaporated  at  or  below  100°  C.,  and  converted  afterwards 
into  steam  at  a  high  temperature,  I  recommend  first  calculating 
the  heat  required  to  convert  the  water  into  vapor  at  0°  (606.5 
Cal.)  and  then  treating  the  vapor  afterwards  as  true  gas,  by 
the  formulas  of  Mallard  and  Le  Chatelier.  This  puts  water 
vapor  at  once,  from  the  beginning,  on  the  same  footing  as  all 
the  other  gases,  and  greatly  simplifies  all  subsequent  calcula- 
tions, without  introducing  any  unnecessary  errors. 


120  METALLURGICAL  CALCULATIONS. 

Latent  heat  of  fusion  =  80  Cal.  (Bunsen). 

Latent  heat  of  vaporization  =  606.5  Cal.  at  0°  (Regnault). 

=  600.0  Cal.  at  10°. 
=  537.0  Cal.  at  100°  (to  vapor  at 

760mm). 
Total    heat    in    saturated 

steam  =  606.5 +  0.305t  (Regnault). 

Most  metallurgical  problems  which  have  to  do  with  convert- 
ing moisture  in  fuel,  ores,  etc.,  into  vapor,  are  concerned  with 
the  evaporation  of  the  water  far  below  100°  by  means  of  dry 
gases,  into  which  the  water  vapor  enters  at  a  partial  pressure, 
which  is  only  a  small  fraction  of  atmospheric  pressure.  The 
problem  is,  in  these  cases,  to  find  how  much  heat  is  necessary 
to  convert  the  water  into  vapor  at  the  low  temperature  corre- 
sponding to  such  low  pressure.  By  finding  the  temperature 
corresponding  to  said  partial  pressure,  as  a  maximum  tension, 
applying  Regnault 's  formula,  we  get  the  heat  absorbed  in  pro- 
ducing the  vapor  at  that  temperature.  We  can  then  add  to 
this  the  heat  required  to  raise  the  vapor  up  to  the  higher  tem- 
perature at  constant  pressure,  using  Mallard  and  Le  Chatelier's 
formulae. 

Example:  Wet  peat  is  dried  in  a  kiln  by  hot  air,  the  issuing 
air  being  at  50°  C.  and  the  tension  of  the  moisture  in  it  25 
millimeters.  How  much  heat  is  absorbed  in  latent  heat,  and 
how  much  as  sensible  heat  per  1  kilogram  of  water  evaporated? 

Solution:  The  tension,  25  millimeters,  is  the  maximum 
tension  of  aqueous  vapor  at  26°  (Tables).  To  change  1  kilo- 
gram of  water  at  zero  to  saturated  vapor  at  26°  requires,  ac- 
cording to  Regnault 's  formula,  606.5+0.305  (26)  =  614  Calories. 
(The  real  latent  heat  of  vaporization  at  26°  is,  therefore,  614 — 26 
=  588  Calories.)  The  heat  required  to  evaporate  1  kilogram  of 
water  under  the  conditions  of  the  peat  kiln  is  therefore: 

Heat  to  produce  vapor  at  25  m.m.  (26°)  =  614.0  Calories. 

Heat  to  raise  vapor  from  26°  to  50°  (constant 
pressure)  =  [0.42  +  0.000185   (50  +  26)]X 

(50—26)  =   JKX4 

Total  =  624.4 

In  most  cases  of  drying  or  evaporation  the  results  will  be 
sufficiently  accurate  by  assuming  the  water  first  evaporated  at 


THERMOPHYSICS  t)F  CHEMICAL  COMPOUNDS.         121 


0°,  with  latent  heat  of  evaporation  606.5  Calories,  and  then 
raised  as  gas  to  the  end  temperature.  The  difference  between 
this  method  and  the  above  will  be  small,  where  the  temperature 
of  the  issuing  gas  is  below  100°  and  the  proportion  of  vapor  in 
it  small;  where  the  temperature  of  issuing  gas  is  high  and  the 
amount  of  vapor  in  it  large,  the  more  exact  method  should  be 
used.  To  facilitate  this,  the  maximum  tension  of  aqueous 
vapor  for  temperatures  up  to  100°  is  here  given: 


Max.  Tension. 

Max.  Tension. 

Temperature  C°. 

mm.  of  Hg. 

Temperature  C°. 

mm.  of  Hg. 

0° 

4.6 

50° 

92.0 

5° 

6.5 

55° 

117.5 

10° 

9.1 

60° 

148.9 

15° 

12.7 

65° 

187.1 

20° 

17.4 

70° 

233.3 

25° 

23.5 

75° 

288.8 

30° 

31.5 

80° 

354.9 

35° 

41.8 

85° 

433.2 

40° 

54.9 

90° 

525.5 

45° 

71.4 

95° 

633.7 

50° 

92.0 

100° 

760.0 

Beryllium  Oxide,  Be2O8 

Sm  (0°— 100°) 
Boric  Oxide,  B2O8: 

Sm  (16°— 98°) 
Carbonous  Oxide,  CO : 

Sm  (0°  to  t°)   1  m3  up  to 
2000° 

1  m3  2000°— 4000° 
1  ft3  up  to  2000° 
1  ft3  2000°— 4000° 
1  kilo,  up  to  2000° 
1  kilo.  2000°— 4000° 
1  Ib.  up  to  2000° 
1  Ib.  2000°— 4000° 
1  ft.3  up  to  3600°  F. 
1  ft.  3600—7200°  F. 
1  Ib.  up  to  3600°  F. 
1  Ib.  3600°— 7200°  F. 


0.2471  (Nilson  and  Pettersson). 
0.2374  (Regnault). 


0.303 +  0.000027t  (Mallard  and 
Le  Chatelier). 

0.2575  +  0.000072t  (Berthelot). 
0.0189  +  0.0000017tlb.  Cal. 
0.0161 +0.0000045t  Ib.  Cal. 
0.2405 +  0.0000214t  Cal. 
0.2044 +  0.000057t  Cal. 
0.2405 +  0.0000214tlb.  Cal. 
0.2044 +  0.000057t  Ib.  Cal. 
0.0189  +  0.0000009*  B.  T.  U. 
0.0161  +  0.0000025*  B.  T.  U. 
0.2405  +  0.0000119*  B.  T.  U. 
0.2044  +  0.000032*  B.  T.  U. 


122  METALLURGICAL  CALCULATIONS. 

Carbonic  Oxide,  CO2: 

There  has  been  much  doubt  about  the  specific  heat  of  this 
gas,  and  several  experimenters  have  given  it  particular  atten- 
tion. Direct  combustion  of  CO  to  CO2  in  air  has  recently 
given  Mallard  and  Le  Chatelier  a  directly-observed  temperature 
of  2050°,  while  the  formula,  which  has  so  far  been  considered 
by  us  as  the  most  reliable  (Sm  =  0.37  +  0.00027t),  leads  to 
1947°.  After  renewed  consideration  of  the  whole  subject  the 
writer  considers  the  best  values  those  given  below,  because  by 
accepting  these  they  will  agree  with  the  observed  temperature 
of  combustion,  2050°.  We  will  hereafter  use  these  values 
instead  of  the*  one  just  above.  Above  2000°  the  values  of 
Berthelot  and  Vielle  are  the  only  ones  to  be  used. 
Sm  (0°  to  t°)  1  m3  up  to 

2050°  «  0.37  +  0.00022t    (calculated   by 

Richards) . 

1  m3  2000°— 4000°  =  0.815  +  0.0000675t  (Vielle). 

1  ft3  up  to  2050°  =  0.023  +  0.000014t  lb.  Cal. 

1  ft3  2000°— 4000°  =  0.051  +  0.0000042t  lb.  Cal. 

1  kilo,  up  to  2050°  =  0.19 +  0.0001  It  Cal. 

1  kilo.  2000°— 4000°  =  0.42  +  0.000034t  Cal. 

1  lb.  up  to  2050°  =  0.19  +  O.OOOllt  lb.  Cal. 

1  lb.  2000°— 4000°  =  0.41  +  0.000034t  lb.  Cal. 

1  ft3  up  to  3700°  F.  =  0.023  +  0.000008*  B.  T.  U. 

1  ft3  36000  —7200°  F.  =  0.051  +  0.0000023*  B.  T.  U. 

1  lb.  up  to  3700°  F.  =  0.19  +  0.00006*  B.  T.  U. 

1  lb.  3600°— 7200°  F.  =  0.41  +  0.000019*  B.  T.  U. 

In  all  the  above  formulae,  Q  from  t°  to  0°  is  equal  to  Sm 
multiplied  by  t;  e.g.,  Q   (0°  to  t°)   1  m3  to  2050°  =  0.37t  + 
0.00022t2. 

Nitrogen  Peroxide,  NO2: 

Sm  (27°— 280°)  1  m3  =  1.35  (Berthelot  and  Ogier). 

1  kilo.  =  0.65  Cal. 

Magnessium  Oxide,  MgO : 

Sm  (24°— 100°)  =  0.2440  (Regnault). 

Sm  (0°— 1°)  =  0.2420 +  0.000016t  (assumed). 

Mg  (OH2)  Sm  (19°— 50°)  =  0.312  (Kopp). 


THERMOPHYS1CS  OF  CHEMICAL  COMPOUNDS.         123 

Aluminium  Oxide,  A12O3  (alumina,  corundum): 

Sm(0°tot°)  =  0.2081  +  0.0000876t   (constant  by 

Regiiault,    coefficient    of    t    t)y 
Richards,  on  a  corundum  crystal 
tested  up  to  1200°). 
Melting  point  =  about  2050°. 

Latent  heat  of  fusion        =  2.  IT  =  4,878  Cal.  for  A12O3. 

4,878 
=  -y^  =  48  Cal-  Per  Wo, 

Heat  in  solid  at  2050°        -  795  Cal.  by  formula. 

Total  heat  in  liquid  to  0°  =  843  Cal. 

Specific  heat,  liquid  =  0.567  =  specific  heat  of  solid  at 

M.  P. 

Silicon  Oxide,  SiO2  (silicon,  quartz): 

Sm(0°tot°)  =  0.1833  +  0.000077t    (constant    by 

Regnault,  coefficient  of  t  by 
Richards,  on  clear  quartz  up  to 
1200°). 

Latent  heat  of  fusion  (at 

1750°)  =  135  Cal.  (Vogt). 

Melting  point  =  1900°  (Boudouard). 

Latent  heat  of  fusion        =  2.  IT  =  4,563  Cal.  for  SiO2. 


=  76.1  Cal.  per  kilo. 

=  135  Cal.  (Vogt). 

Heat  in  solid  at  1900°       =  626  Cal.  by  formula. 
Total  heat  in  liquid  to  0°  =  761  Cal. 

Specific  heat,  liquid  =  0.476  =  specific  heat  of  solid  at 

M.  P. 
Sulphurous  Oxide,  SO2: 

Sm  (16°—  202°)  1  m»         =  0.4447  (Regnault). 
1  kilo.      =  0.1544. 

If  we  assume  that  the  molecular  specific  heat  of  SO2  is  8 
Calories  (from  the  rule  that  the  molecular  specific  heat  of  a 
gas  at  constant  pressure  is  5-fn,  where  n  is  the  number  of 
atoms  in  the  molecule),  we  would  have 

S  (at  0°)  1  m3  -  =  °'36  CaL 


124  METALLURGICAL  CALCULATIONS. 

Combining  this  with  Regnault's  value  of  Sm,  we  would  get  the 
formula 

Sm  (0°  to  t°)  1m3  =  0.36  +  0.0003t 

1  kilo.          =  0.125  +  O.OOOlt 

These  values  are  probably  accurate  enough  for  furnace 
calculations,  and  are  very  useful  in  pyritic  smelting  and  the 
roasting  of  sulphide  ores.  While  it  is  always  desirable  to  have 
direct  determinations  of  such  important  quantities,  yet  when 
they  have  never  been  made  it  is  allowable  to  work  out  and 
use  the  most  probable  values. 

Sulphuric  Oxide,  SO3: 

The  specific  heat  of  this  important  gas  has  not  been  measured. 
As  an  approximation  we  may  assume 

S  (atO°)  per  molecule        =  9.0  Cal.  (assumed), 
per  m3  =  0.405  Cal. 

per  kilogram       =  0. 11  Cal. 

For  getting  an  approximation  to  Sm  we  may  assume  the 
coefficient  of  t  the  same  as  in  NH3,  which  contains  the  same 
number  of  atoms  and  is  of  analogous  formula,  and  we  then 
have 

Sm  (0°  to  t°)  per  molecule  =  9.0     +0.0036t 
perm3  =  0.405 +  0.00017t 

per  kilogram      =  0. 100  +  0.00004t 

Calcium  Oxide,  CaO  (lime): 

This  important  datum  has  not  been  determined,  but  since 
it  is  so  closely  analogous  to  MgO,  an  approximation  may  be 
obtained  by  assuming  that  the  molecular  specific  heats  of  the 
two  compounds  are  alike.  That  for  MgO  is  0.2440x40  =  9.76 
Calories.  That  for  CaO  would  therefore  be 

Q  7fi 

Sm  (24°— 100°)  =  ^~  =  0.1743 

ou 

and  assuming  a  usual  coefficient  of  t;  for  compounds  of  this 
type: 

Sm  (0°— 1°)  =  0.1715  +  000007t 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        125 


Titanic  Oxide,  TiO2: 
S  (up  to  200°) 
Q  (0°  to  200°) 
Q  (0°  to  t°)  for  t  over  200° 

Chromium  Oxide,  Cr208: 
Sm  (10°— 99°) 

Manganese  Oxides: 
MnO  Sm  (13°— 98°) 
Mn203  Sm  (15°— 99°) 

(Manganite)  Mn203H20 
Sm  (21°— 52°) 

(Pyrolusite)  MnO2 
Sm  (17°— 48°) 

Iron  Oxides.  Fe2O3: 
Sm  (0°— 1°) 

Fe304  Sm    0°— 1°) 


0.1790  (Nilson  and  Pettersson). 

35.8  Cal. 

35.8  +  0.1790  (t— 200) +  .000055 

(t— 200)2. 

=  0.1796  (Regnault). 

=  0.1570  (Regnault). 
=  0.1620  (Oeberg). 

=  0.1760  (Kopp). 
=  0.1590  (Kopp). 

-  0.1456  + 0.0001 88t  (Regnault  and 

Richards). 

=  0.1447  +  0.000188t  (Regnault  and 

Richards) . 


0.1588  (Regnault) 


0.1110  (Oeberg). 
0.1420  (Regnault). 


Nickel  Oxide,  NiO: 
Sm  (13°— 98°) 

Copper  Oxides-. 

Cu20  Sm  (19°— 51°) 
CuO  Sm  (12°— 98°) 

Zinc  Oxide,  ZnO: 

Sm  (0°— 1°)  (tot  =  1000°)  =  0.1212  +  0.0000315t  (Richards). 

Arsenious  Oxide,  As2Os: 
Sm  (13°— 97°) 

Zirconium  Oxide,  ZrO2: 
Sm  (0°— 100°) 

Columbic  Oxide,  Cb205 
Sm    0°— 1° 


0.1276  (Regnault). 


=  0.1076  (Nilson  and  Pettersson). 


Molybdic  Oxide,  MoOs: 
Sm  (21°— 52°) 


0.1037  +  0.000070t(Kruss  and  Nil- 
son). 

0.1540  (Kopp). 


126 


METALLURGICAL  CALCULATIONS. 


Tin  Oxide,  SnO2: 

Sm  (16°— 98°) 
Sm  (0— t) 

Antimonious  Oxide,  Sb2O3: 

Sm  (18°— 100°) 
Tungstic  Oxide,  WO3: 

Sm  (8°— 98°) 
Mercuric  Oxide,  HgO: 

Sm  (5°— 98°) 
Lead  Oxide,  PbO: 

Sm  (22°— 98°) 
Bismuth  Oxide,  Bi2O3: 

Sm  (20°— 98°) 
Thoric  Oxide,  Th2O3: 

Sm  (0°— 100°) 


0.0936  (Regnault). 

0.1050+  0.000006t  (Richards). 


=  0.0927  (Neumann). 


=  0.0798  (Regnault). 


=  0.0518  (Regnault). 


=  0.0512  (Regnault). 


0.0605  (Regnault). 


=  0.0548  (Nilson  and  Pettersson). 


For  those  metallic  oxides  whose  specific  heat  has  not  been 
determined,  an  approximation  to  the  specific  heat  can  be 
found  by  assuming  that  the  metallic  atoms  in  a  molecule  of 
oxide  have  each  a  specific  heat  6.4  Calories,  and  each  atom  of 
oxygen  in  an  oxide  molecule  has  an  atomic  specific  heat  of  3.6. 
As  a  rough  approximation,  we  can  further  assume  that  the 
mean  specific  heat  from  0°  to  t°  increases  0.04  per  cent,  for 
each  degree  of  temperature,  so  that  the  above  calculated  spe- 
cific heat  being  from  0°  to  100°,  we  can  figure  out  S  at  zero 
and  the  rate  of  increase  of  S  or  Sm  with  t. 

Example:  What  is  the  most  probable  value  of  the  specific 
heat  of  CaO?  Since  molecule  weight  is  56,  and  the  molecule 
may  be  assumed  to  have  a  molecular  specific  heat  of  6.4  +  3.6 
=  10.0  Calories,  the  mean  specific  heat  per  kilogram  (0°— 100°) 
is  lO.O-i-56  =  0.1786.  S  at  zero  will  then  be  0.1786-5-1.04  = 
0.1717,  and  we  would  have  the  formulae: 

Sm  =  0.1717  +  0.00007U 
S  =  0.1717  +0.000142t 
Q  =  0.1717t  +  0.000071t3 


THERMOPHYS1CS  OF  CHEMICAL  COMPOUNDS.         127 

Problem  13. 

The  Jacobs  process  of  producing  artificial  emery  consists  in 
melting  down  impure  bauxite  in  an  electric  furnace,  and  letting 
the  mass  cool.  Assume  the  bauxite  to  be  calcined  before  use, 
and  to  contain  Per  cent. 

Alumina . . , 88 

Ferric  oxide  (Fe2O3). 5 

Silica  (SiO2) 5 

Titanic  oxide  (TiO2) 2 

Required : 

(1)  The  heat  necessary  to  just  melt  a  metric  ton  of  this 
material  at  2000°. 

(2)  The  net  electric  power  theoretically  required. 

(3)  The  total  electric  power  actually  required. 

Solution : 

(1)  Because  of  the  presence  of  the  impurities  the  alumina 
melts  easier  than  it  otherwise  would.  We  can,  therefore, 
calculate  the  heat,  in  the  alumina  melted  at  2000°,  and  the 
heat  in  the  others  melted  at  their  proper  melting  points,  because 
they  absorb  their  latent  heats  of  fusion  at  the  lower  tempera- 
ture. Take  1000  kilos,  of  material. 

Heat  in  1  kilo,  of  alumina: 

Q  (0°— 2000°)  =  0.2081  (2000) +0.0000876  (2000)2       =767  Cal. 
L.H.F.  at  2000°   =   2.1   (2000  +  273) -^  102  (mol.  wt. 

A12O3)  =_47     " 

Total      =814     " 
Heat  in  1  kilo,  of  ferric  oxide 

Q  (0°— 1600°)  =  0.1456  (1600) +0.000188  (1600)2  (assumed) 

=  713  " 

L.H.F.  at  1600°  =  2.1  (1600  +  273) -*•  160  =25  " 

Heat  in  liquid  =  (2000— 1600)  X  0.75  =300  " 

Total      =1038  " 
Heat  in  1  kilo,  of  silica: 

Q  (0°— 1900°)  =  0.1833  (1900) +0.000077  (1900)2  =626  " 

L.H.F.  (Voigt)  =135  " 

Q  (1900°— 2000°)  liquid  =  0.476X100  =   48  " 

Total  =809  '* 


128  METALLURGICAL  CALCULATIONS. 

Heat  in  1  kilo,  of  titanic  oxide  : 
Q  (0°—  2000°)  =  35.8+0.1790  (1800)4-0.000055 

(1800)2  =  536  Cal. 

L.H.F.  =  2.  IT  =  2.1  (2000  +  273)4-80  =    59    « 

Total      =  595    " 
Total  heat  to  fuse  1000  kilos.  : 

Heat  in  alumina  880X814  »  716,320  Cal. 

Heat  in  ferric  oxide  50X1038  «=    51,900     " 

Heat  in  silica  50X809  =    40,450     * 

Heat  in  titanic  oxide          20X595  =     11.900     " 

Total  =  820,570    " 

(2)  One  kilowatt  equals  859  Calories  per  hour. 

820,570 
Kilowatt-hours  absorbed  by  melt  —  —  zZn  —  =  955.3  kw-hr. 


(3)  Assume  an  efficiency  of  50  per  cent.,  requires  1911  kw-hrs. 
If  the  dynamos  work  at  95  per  cent.,  efficiency,  the  working 
power  of  the  water  turbine,  gas  engines  or  steam  engines  would 

1Q11 

bei^-i  =  2000kw.hr. 
u  .  y  o 

Problem  14. 

The  ferric  oxide  charged  into  a  blast  furnace  is  reduced  at 
900°  C.  by  carbonous  oxide  gas,  CO,  according  to  the  reaction: 
Fe2O3  +  9CO+  (17.15N2)  =  Fe2  +  3C02  +  6CO  +  (17.15N2). 

Required: 

(1)  The  thermal  value  of  the  reaction  at  900°. 

(2)  What  would  be  the  temperature  of  the  resulting  products 
after  the  reaction  had  occurred? 

Solution  : 

(1)  The  6CO  and  17.15N2  occurring  on  both  sides  of  the  equa- 
tion need  not  be  considered.  The  heat  value  of  the  equation 
at  ordinary  temperatures  would  therefore  be: 

Decomposition  of  Fe203  =  —  195,600  Cal. 

Decomposition  of  3CO  =  —  87,480     " 

Formation  of  3CO2  •  +291,600     « 

Thermal  value  of  reaction  «  +     8,520     " 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.         129 

At  900°,  the  value  is  changed  as  follows: 
Added: 

Heat  in  Fe2O3  (160  kilos.)  at  900°  =  160  [0.1456 

(900)-f0.000188(900)2]  =  +  45,200  Cal. 

Heat*  in   SCO    (=  66.66m8)   at   900°  =  66.66 

[0.303(900)+  0.000027  (900)2]  =  +   19,640    " 

Total  to  be  added  =  +   64,840    " 

Subtracted: 

Heat  in  Fe2  (112  kilos.)  at  900°  =  112 

X(0.218t—  39)  =        17,585    " 

Heat    in    3C02    (66.66   m3)    at    900°  =  66.66 

[0.37  (900)+  0.00022  (900)2]  =        34,080    " 

Total  to  be  subtracted  51,665    " 

Heat  of   reaction   at   900°  =  8,520  +  64,840  — 

51,665  =  +   21,695    " 

(2)  The  above  heat  would  serve  to  raise  the  temperature  of 
all  the  products,  viz.:  Fe2,  3CO2,  6CO,  17.15N2  (from  blast). 
Their  thermal  capacity,  900°  to  t,  is: 

Fe2     =  112  [0:218(t—  900)] 
3CO2  =  66.66    [0.37  (t—  900)+  0.00022  (t2—9002)] 
6CO    =  133.33  [0.303(t—  900)  +0.000027  (t2—  9002)] 
17.15N2      =  17.15    [  "  ] 

Sum  =  0.018517t2  +  94.68t-  100.383  =  21,695  Calories. 
Whence  t  =  1065°. 

It  follows  from  the  preceding  discussion  that  the  reduction  of 
iron  oxide  by  CO,  at  about  900°,  is  an  exothermic  reaction, 
heat  being  evolved,  and  that  the  reaction  proceeds  to  a  finish 
when  once  started.  The  relations  for  FeO  would  be  probably 
similar,  but  we  do  not  have  the  specific  heat  of  FeO  to  use  in 
making  the  exact  calculations. 

Problem  16. 

If  Fe2O3  charged  into  a  blast  furnace  were  reduced  by  solid 
carbon,  at  900°  C.,  by  the  reaction 


4Fe2Os  +  9C  =  3CO2  +  6CO  +  4Fe>. 
Required: 

(I)  The  thermal  value  of  the  reaction  at  900 


130  METALLURGICAL  CALCULATIONS. 

(2)  The  temperature  of  the  resultant  products  after  the  reac- 
tion. 

Solution : 

(1)  Value  of  the  reaction  at  zero  =-  4(195,600)  Cal. 

+  3(97,200) 
+  6(29,160) 

Add:  •    315,840 

Heat  in  4Fe2O3  at  900°  =  +  180,800      " 

Heat  in  9C  at  900°  =  9(12)  [0.2142(900)  + 

0.000166(9002)]  =  +     35,340      " 

Subtract: 

Heat  in  4Fe2  at  900°  =  —  70,340  " 

Heat  in  3CO2  at  900°  =  —  34,080  " 
Heat  in  6CO  at  900°  =  6  (22.22)  [0.303  (900) 

+  0.000027  (9002)]  =  —  39,275  " 

Algebraic  sum  =  —  243,395  " 

Absorbed  per  molecule  of  Fe2O3  =  —  60,850  " 

This  reaction  is  therefore  strongly  endothermic,  and  therefore 
tends  to  check  itself  constantly  by  the  resultant  cooling  effect. 

(2)  If  the  Fe203  and  C  were  at  900°  to  start  with,  the  pro- 
ducts would  be  below  900°  after  the  reaction,  since  they  would 
have  to  furnish  the  deficit  of  heat. 

The  products  would  give  out  in  cooling  from  900°  to  t°. 

4Fe2   =    4(112)   [0.218(900— t)] 

SCO2  =  66.66       [0.37(900— t) +0.00022  (9002—t2)] 

6CO    =  133.33     [0.303(900— 1)+  0.000027  (9002— t2)] 

Sum  =  0.0183t2— -162.73t  +  161,250,=  243,395  Cal. 

Whence  t  =  —  537°. 

This  result  is,  of  course,  due  to  the  hypothetical  assumption 
that  this  endothermic  reaction  would  go  on  until  completed 
without  checking  itself.  The  impossible  result  obtained  means 
that  if  this  single  reaction  really  goes  on,  it  will  soon  check 
itself  on  account  of  the  decrease  of  temperature. 


There  are  a  few  other  compounds  than  those  previously 
mentioned  whose  thermophysics  has  been  investigated,  but 
their  number  is  in  reality  infinitesimal  compared  with  the  num- 
ber of  those  which  have  not  been  touched.  There  are  many 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.         131 

compounds  of  great  importance  in  metallurgy  whose  specific 
heats  are  not  known,  to  say  nothing  of  their  latent  heats  of 
fusion,  etc.  The  wide  introduction  of  the  electric  furnace  has 
rendered  desirable  the  latent  heats  of  vaporization  and  specific 
heats  at  high  temperatures,  but  these  are  altogether  lacking; 
we  can  only  estimate  their  values.  It  is  to  be  hoped  that  many 
metallurgical  laboratories  may  be  incited  to  take  up  this  very 
neglected  field,  and  thus  bring  forward  numerical  data  which 
would  be  of  the  greatest  theoretical  and  practical  value.  One 
example  of  such  work,  which  deserves  special  commendation, 
is  the  quite  recent  publication  of  Prof.  J.  H.  L.  Vogt,  of  Chris- 
tiania,  on  the  "  Silikatschmelzlosungen,"  i.e.,  on  "  Melted 
Silicate  Solutions,"  in  which  determinations  of  the  melting 
points  and  latent  heats  of  fusion  of  many  simple  and  complex 
silicates  are  given  for  the  first  time ;  we  venture  to  predict 
that  the  data  and  conclusions  of  Prof.  Vogt  will  be  of  great 
value  to  the  physicist,  chemist,  metallurgist,  and  geologist,  in 
both  a  practical  as  well  as  a  theoretical  sense. 

In  collating  the  remaining  available  data,  it  is  rather  difficult 
to  decide  as  to  wha.t  has  immediate  metallurgical  interest  and 
what  has  not.  Electrometallurgy,  in  particular,  is  busying 
itself  with  the  treatment  and  decomposition  of  so  many  com- 
pounds heretofore  considered  outside  of  the  metallurgist's 
sphere  of  interest,  that  almost  all  of  the  common  chemical 
compounds  now  have  either  a  present  or  a  prospective  interest 
to  the  metallurgist. 


THERMOPHYSICS  OF  CHLORIDES 


Substance 
HC1  (gas) 

Tem- 
per a  tur  e 

22-214° 

L.  H. 
Specific    Heat         Fusion 

0  1867 

L.  H. 

Va  port-    A  uthority 
zalion 

Regnault 

HC1  (liquid) 

-83 

0.3067(lm3) 

97  .  5  Elliott  & 

RbCl  (fused) 

16-45 

0  112               

Mclntosh 
.  Kopp 

LiCl  (fused) 

13-97 

0  2821              

Regnault 

KC1 
KC1 

14-99 

77-2 

0.1730             
86.0 

Regnault 
Plato 

NaCl 

15-98 

0  2140 

Regnault 

NaCl 

804 

123  .  5 

Plato 

NH4C1 

23-100 

0.3908 

.   Neumann 

132 


METALLURGICAL  CALCULATIONS. 


Substance 

Tem- 
perature 

Specific  Heat 

L.  H.      L.  H. 

Fusion       Vapor  'i-    Authority 

zation 

NH4C1 

350° 

709  Marignac 

BaCl2  (fused) 

14-98 

0.0896 

Regnault 

BaCl2  (fused) 

959 

27  .  8       Plato 

SrCl2 

13-98 

0.1199 

Regnault 

SrCl2 

872 

25.6       Plato 

CaCl2  (fused) 

23-99 

0.1642 

Regnault 

CaCl2  (fused) 

774 

54.6       Plato 

MgCU  (fused) 

24-100 

0.1946 

Regnault 

A1C13 

-22-15 

0.188 

Band 

TiCl4  (solid) 

13-99 

0.1881 

Regnault 

TiCl4  (gas) 

163-271 

0.1290 

Regnault 

SiCl4  (liquid) 

10-15 

0.1904 

Regnault 

SiCl4  (gas) 

90-234 

0.1322 

Regnault 

SiCl4  (gas) 

90-234 

1.0113(lm3) 

Regnault 

SiCl4  (liquid) 

? 

37.3     Ogier 

CC14  (liquid) 

20° 

0.207 

Timofejew 

CC14  (liquid) 

0° 

52  .  0    Regnault 

MnCl2 

15-98  (?) 

0.1425 

Regnault 

MnCl2  (liquid) 

? 



49.  4(?)  Ogier. 

ZnCl2  (fused) 

21-99 

0.1362 

Regnault. 

/      0  0525 

T1C1  (solid) 
TJC1  (at  M.  P.) 

0-t 
427° 

I  +0.000007t 
0.0585 

Russell, 
Goodwin  & 

T1C1  (solid) 

427° 

15.6     

T1C1  (liquid) 

427-530 

0.0590 

CrCl2 

? 

0.1430 

Kopp 

PC13  (solid) 

11-98 

0.2092 

Regnault 

PC13  (liquid) 

78.5 

51.42  Andrews 

PCI,  (gas) 

111-246 

0.135 

Regnault 

AsCl,  (liquid) 

14-98 

0.1760 

Regnault 

AsCls  (liquid) 



69  .  74  Regnault 

AsCl3  (gas) 

159-268 

0.1122 

Regnault 

SnCl2  (fused) 

20-99 

0.1016 

Regnault 

SnCl4  (liquid) 

10-15 

0.1904 

Regnault 

SnCl4  (liquid) 

112 

46.84  Regnault 

SnCl4  (gas) 

149-273 

0.0939 

Regnault 

HgCl  (solid) 

7-99 

0.0521 

Regnault 

HgCl2  (solid) 

13-98 

0.0689 

Regnault 

CuCl  (solid) 

17-98 

0.1383 

Regnault 

PbCl2  (solid) 
PbCU  (at  M.  P 

0-t 
)          498 

/      0.0630 
I  +0.00002U 
0.0839 

Lindner, 
Goodwin  & 

PbCl2  (solid) 

498 

18.5       

PbCl2  (liquid) 

>       498 

0.1035 

Ehrhardt 

AgCl  (fused) 

0-t 

f      0.0897 
I  +0.0000125t 

f  Regnault 
I  Goodwin, 

THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        133 


Substance 

Tem- 
peratur 

L.  H. 

Specific  Heat          Fusion 

L.  H. 

Vapor  i-        Authority 

zation 

AgCl  (at  M.  P.) 

455° 

0.1011              

Robertson 

AgCl  (solid) 

455 

30.5 

& 

AgCl  (liquid) 

455-533 

0.129               

Kalmus 

THERMOPHYSICS  OF  BROMIDES 

HBr  (gas) 

11-100 

0.0820             

Strecker 

HBr  (gas) 

11-100 

0.2989  (perm*)    

Strecker 

HBr  (liquid) 

-70 

48.68  Estreigher    & 

Schnerr 

KBr  (fused) 

16-98 

0.1132             

Regnault 

NaBr 

0-96 

0.1170             

Koref 

AlBr3 

93° 

10.47 

Kablukow 

SbBr3 

95 

9.76 

Telloczko 

AsBr3 

31 

8.93 

Telloczko 

SnBr4 

30 

6.26 

Telloczko 

TIBr 

460 

.......           12.7 

Goodwin  & 

Kalmus 

PbBr2  (fused) 

0-t 

f      0.0526             
\+0.000006t        

Goodwin  & 
Kalmus 

PbBr2  (at  M.  P.) 

488 

0.0585             

from  formula 

PbBr2  (at  M.  P.) 

488 

9.9 

G.  &  K. 

PbBr2  (at  M.  P.) 

488 

12.34 

Ehrhardt 

PbBr2  (liquid) 

488-587 

0.0780             

G.  &  K. 

0  07^6 

Regnault, 

AgBr  (fused) 

0-t 

+0.0000025t      

Goodwin  & 

Kalmus. 

AgBr  (at  M.  P.) 

430 

0.0757              

Calculated 

AgBr  (at  M.  P.) 

430 

:  12.6 

G  &  K. 

AgBr  (liquid) 

430-563 

0.0760             

G.  &  K. 

THERMOPHYSICS  OF  IODIDES 

HI  (gas) 

21-100 

0.0550             ..... 

Strecker 

HI  (gas) 

21-100 

0.3168  (perm3)    

Strecker 

HI  (liquid) 

-37 



/  Elliott  & 
o5.00   <  -_  T    ,     , 
I  Mclntosh 

KI  (solid) 

13-48 

0.0766             

Nernst 

Nal  (solid) 

16-99 

0.0868             

Regnault 

Cul  (solid) 

18-99 

0.0687             

Regnault 

Hg,I  (red) 

0-100 

0.0406             

Guinchan' 

Hg2I  (yellow) 

0-247 

0.0446             

.  .  .^.  .   Guinchant 

Hg2I  (liquid) 

250-327 

0.0554             

Guinchant 

Hgl  (solid) 

17-99 

0.0395             

Regnault 

HgI2  (solid) 

18-99 

0.0420             

Regnault 

Hglj  (at  M.  P.) 

250 

9.79 

Guinchant 

134 


METALLURGICAL  CALCULATIONS. 


Substance 


Tem- 
perature 


Specific  Heat 


L.  H. 

Fusion 


L.  H. 

Vapori- 
zation 


Authority 


PbI2  (solid)                 0-375°  0.0430  Ehrhardt 

PbI2  (at  M.  P.)            375  11 . 50     Ehrhardt 

PbI2  (liquid)      >         375  0.0645  Ehrhardt 

Agl  (solid)                15-264  0.0577  Bellati  & 

Romanese 

PbI2.AgI                    10-124  0.0476  Bellati  & 

Romanese 

THERMOPHYSICS  OF  FLUORIDES 

HP  (liquid)  .                    ?  ....       360    Guntz 

KP                          -76-0  0.1930 Koref 

KP  (M.  P.)                   860  .......  108       Plato 

NaP                           15-53  0.2675  Band 

3NaP.AlF3                16-99  0.2522 Oeberg 

(Cryolite) 

CaF2                           15-99  0.2154  Regnault 

A1F3                            15-53  0.2294 Band 

A1F3.7H2O                 15-53  0.342  Band 

PbF2                            0-34  0.0722  Schottky 

THERMOPHYSICS  OF  SULFIDES 

H2S  (gas)                  20-206  0 . 2451  Regnault 

H2S  (gas)                   20-206  0.3750(lm3)  Regnault 

H2S  (gas)                    0-t  (Im3)0.34 Richards 

+0.00015t 

CS2  (liquid)               14-29  0.2468  Person 

CS2  (gas)                    86-190  0. 1596  .  .  . Regnault 

CS2  (gas)                   86-190  0.5458(lm3)  Regnault 

CS2  (liquid)                   46  83, 8     Wirtz 

MnS                            10-100  0.1392                Sella 

ZnS                             15-98  0. 1230 Regnault 

CdS                             26  0.0908  Russell 

FeS                             17-98  0. 1357  Regnault 

Fe7S8                          20-100  0.1602  .., Regnault 

(Pyrrhotite) 

FeS2                            19-98  0. 1301 Regnault 

FeS2                                565  (Gives  off  S)  Richards 

CoS                             15-98  0. 1251 Regnault 

NiS                             15-98  0 . 1281  Regnault 

MoS2                         20-100  0.1233  Regnault 

(Molybdenite) 

Cu2S                             9-97  0. 1212  Regnault 

(Chalcocite) 

103  Transformation  Point /  Bellati  & 

190  0.1454                                  ,    \Lussana 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS. 


135 


Tem- 

L. H.       L.  H. 

Substance 

perature 

Specific   Heat          Fusion     Vapori- 

Authority 

zation 

CuS 

25° 

0.1243              

Russell 

(Covellite) 

AsS 

20-100 

0.1111              

Naumann 

(Realgar) 

As2S3 

20-100 

0.1132              

Naumann 

(Orpiment) 

(Orpiment)  (gas) 

0-t 

(Im3)0.34      

Richards 

+0.00015t 

Sb2S3 

23-99 

0.0840                           

Regnault 

(Stibnite) 

SnS. 

13-98 

0.0837             

Regnault 

SnS2 

12-95 

0.1193             .....      

Regnault 

PbS 

16-98 

0'.0509 

Regnault 

PbS  (liquid) 

1050 

.......             50         

Richards 

(104  to  0°) 

Bi2S3  (fused) 

11-99 

0.0600             

Regnault 

HgS 

14-98 

0.0512             

Regnault 

Ag2S 

7-98 

0.0746             

Regnault 

Ag2S 

0-t 

0.0685             

Bellati  & 

+0.00005t 

Lussana 

THERMOPHYSICS  OF  COMPOUND  SULFIDES 

3Cu2S.Pe2S3 

10-100 

0.1177             

Sella 

(Bornite) 

Cu2S.Fe2S3 

14-98 

0.1310             

Kopp 

(Chalcopyrite) 

(Chalcopyrite) 

720 

(Gives  off  S)    

Richards 

Cu  Matte 

0-t 

0.2110             

Landis 

(47%  Cu) 

-0.0000366t 

Cu  Matte 

1000° 

30 

Landis 

(47%  Cu) 

Cu  Matte  (liquid) 

1000° 

(204        

Landis 

toO°) 

4Cu2S.Sb2S3 

10-100 

0.0987             .....     

Sella 

(Tetrahedrite) 

2PbS.Cu2S.Sb2S3 

10-100 

0.0730             .....      

Sella 

(Bournonite) 

FeS2.PeAs2 

10-100 

0.1030             .....      

Sella 

(Arsenopyrite) 

CoS2.CoAs2 

15-99 

0.0991             ...'..     

Sella 

(Cobaltite) 

3Ag2S.As2S3 

10-100 

0.0807             ..... 

Sella 

3Ag2S.Sb2S3 

10-100 

0.0757 

Sella 

136 


METALLURGICAL  CALCULATIONS. 


THERMOPHYSICS  OF   SELENIDES  AND  TELLURIDES 


Substance 

Tern-                                               L.  H.        L.  H. 
perature           Specific  Heat          Fusion     Vapori- 

A uthority 

zation 

Cu2Se 

20-200°          0.1047             

j    Bellati  & 

Ag2Se 

37-187           0.0684             

/    Lussana 

THERMOPHYSICS  OF  ARSENIDES   AND  ANTIMONIDES 

PeAs2 

10-100           0.0864             

Sella 

(Lollingite) 

CoAs2 

10-100           0.0830             

Sella 

(Smaltite) 

Cu3As 

10-100           0.0949              

Sella 

(Domeykite) 

Ag3Sb 

10-100           0.0558              

Sella 

(Dyscrasite) 

THERMOPHYSICS  OF  CYANIDES 

K4Fe(CN)6 

1-46             0.2142             

Nernst 

K2Zn(CN)2 

14-46             0.241                

Kopp 

Hg(CN)2(cryst) 

11-46             0.100               

Kopp 

THERMOPHYSICS  OF  HYPOSULFITES 

K2S20, 

20-100           0.197               

Pape 

Na2S2O8 

25-100           0.221               

Pape 

BaS2O3 

17-100           0.163               

Pape 

PbS208 

15-100           0.092               

Pape 

THERM.OPHYSICS  OF  SULFATES 

H2SO4 

10.5             26.0       

Bronsted 

H2SO4 

5-22             0.332               

Cattanes 

H2SO4 

326             122.1 

Person 

K2S04 

15-98            0.1901             

Regnault 

KHSO4 

19-51             0.2440             

Kopp 

Na2SO4 

17-98             0.2312             

Regnault 

Na2SO4.10H2O 

31               51.2       

Cohen 

BaSO4 

10-98             0.1128             

Regnault 

SrS04 

21-99             0.1428             

Regnault 

CaSO4 

13-98             0.1965             

Regnault 

MgS04 

25-100           0.2250             

Pape 

A12K2(SO4)3.24H2O 

15-52             0.3490             

Band 

(NH4)2SO4.12H2O 

13-45             0.3500             

Kopp 

ZnSO4 

22-100           0.1740             

Pape 

MnSO4 

21-100           0.1820             

Pape 

Cr3K2(S04)4.24H20 

19-51             0.3240             

Kopp 

THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        137 


Substance 


Specific  Heat 


Authority 


FeSO4.7H2O               9-16° 

0.3460             .;  

Kopp 

CoSO4.7H2O             15-80 

0.3430             

Kopp 

NiS04                        15-100 

0.1840             

Pape 

CuSO4                       23-100 

0.1840             

Pape 

PbSO4                       20-99 

0.0872             

Regnault 

Hg2S04                       0-34 

0.0624             

Schottky 

THERMOPHYSICS  OF  NITRATES 

HNO3                          -47 

9.54  ..... 

Berthelot 

HNO3                            86 

..  115.1 

Berthelot 

LiNOa                       169-250 

0.387              

Goodwin 

&  Kalmus 

LiNOa                           250 

88.5     

Goodwin 

&  Kalmus 

LiNO3  (liquid)        250-302 

0.390               

Goodwin 

&  Kalmus 

KN03                        13-98 

0.239               

Regnault 

KN03                          334° 

48.9     

Person 

KNO3  (liquid)         350-435 

0.332 

Person 

NaNO3                      14-98 

0.278               

Regnault 

NaNO3                          306 

64.9     

Person 

NaNO3  (liquid)       320-430 

0.413               

Person 

K2Na(NO3)2             15-100 

0.235               

Person 

NH4NO3                   14-31 

0.455               

Kopp 

BaNO3                      13-98 

0.1523             

Regnault 

SrNO3                       17-47 

0.181               

Kopp 

CaNO3.4H2O                42 

33.5     

Pickering 

AgNO3                       16-99 

0.1435 

Regnault 

AgNO,                          209 

17.6     ..... 

Guinchant 

AgNO3  (liquid)       208-281 

0.187               

Guinchant 

PbN03                      17-100 

0.1173             .     

Neumann 

THERMOPHYSICS  OF  CARBONATES 

Pb2CO3  (fused)         18-47 

0.123               .....     

Kopp 

K2CO3                       23-99 

0.2162             

Regnault 

Na2CO3                     16-98 

0.2728             

Regnault 

BaCO3                       11-99 

0.1104             

Regnault 

SrCO3                          8-98 

0.1475             

Regnault 

CaCO3  (calcite)        20-100 

0.2086             

Regnault 

CaCO3  (argonite)      18-99 

0.2085             

Regnault 

CaCO3  (marble)       23-98 

0.2099             

Regnault 

CaMg(CO3)2             20-100 

0.2179             

Regnault 

(Dolomite) 

Mg7Fe2(C03)9           17-100 

0.2270             

Neumann 

(Brown  magnesite) 

138 


METALLURGICAL  CALCULATIONS. 


Substance 

pZaTure          Sp^fic  Heat          Fusion 

L.  H 
Vapor  i-         Authority 
zation 

ZnCO3 

o-t°     I    °-1407  1 
l+o.oooitj 

Lindner 

CuCO3.Cu(OH)2 

15-99             0.1763             

Oeberg 

(Malachite) 

FeCO3 

9-98             0.1935             

Regnault 

PbCOs 

16-47             0.0791             

Kopp 

THERMOPHYSICS  OF  CHROMATES 

K2CrO4 

19-98             0.1851             

Regnault 

K2Cr207 

0-t               0.1803             
+0.00008t 

(  Regnault,  & 
•j   Goodwin  & 
I  Kalmus 

K2Cr2O7 

397              29.8 

Goodwin  &  K. 

K2Cr2O7  (liquid) 

397-484           0.335               

Goodwin  &K. 

Na2CrO4 

23               39.2 

Berthelot 

FeCrO4 

19-50             0.1590             

Kopp 

(Chromite) 

PbCrO4 

19-50             0.0900             

Kopp 

THERMOPHYSICS  OF  BORATES 

K2B204 

16-98             0.2048             

Regnault 

K2B407 

18-99             0.2198             

.*....  Regnault 

Na2B2O4 

17-97             0.2571             

Regnault 

Na2B4O7 

16-98             0.2382 

Regnault 

PbB2O4 

15-98             0.0905             

Regnault 

PbB407 

16-98             0.1141             

Regnault 

THERMOPHYSICS  OF  PHOSPHATES 

H8P04 

18               25.71 

Thomsen 

K4P207. 

17-98            0.1910            

Regnault 

Na4P207 

17-98            0.2283             

Regnault 

CaP2O6 

15-98             0.1992             

Regnault 

3Ca3P2O8.CaF2 

15-99             0.1903             

Oeberg 

(Apatite) 

Pb2P2O7  (fused) 

11-98             0.0821             

Regnault 

Pb3P208 

11-98             0.0798             

Regnault 

Ag3P04 

19-50             0.0898             

Kopp 

THERMOPHYSICS  OF  ARSENATES 

Pb8As208  (fused) 

13-97             0.0728             

Regnault 

KaAsaO6  (fused) 

17-99            0.1563             

Regnault 

THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        139 
THERMOPHYSICS  OF  CHLORATES  AND  PER-CHLORATES 


Substance 


Tem- 
perature 


Specific  Heat 


7     f7 

Vapor  i- 


Authority 


KC1O3  (fused)  16-98°  0.2096  Regnault 

KC1O4  14-45  0.190  Kopp 

NaClOa  (fused)       184-255  0 . 320  Goodwin 

&  Kalmus 
NaClO3  (fused)          255  49 . 6     Goodwin 

&  Kalmus 
NaClO3  (liquid)      255-299  0.325  Goodwin 

&  Kalmus 

THERMOPHYSICS  OF  ALUMINATES 

MgAl2O4  15-47  0.1940  Kopp 

(Spinel) 

BeAl2O4  0-100  0.2004  Nilson  & 

(Chrysoberyl)  Petersson 

THERMOPHYSICS  OF  TITANATES 

FeTiOa  15-50  0.177  Kopp 

THERMOPHYSICS  OF  MOLYBDATES 

PbMoO4  15-50  0.083  Kopp 

(Wulfenite) 

THERMOPHYSICS  OF  TUNGSTATES 

CaWO4  15-50  0.097  Kopp 

(Scheelite) 

Fe(Mn)WO4  15-50  0.098  Kopp 

(Wolframite) 

THERMOPHYSICS  OF  MANGANATES 

KMnO4  15-50  0.179  .....     Kopp 

THERMOPHYSICS  OF  SILICATES 

Mg2SiO4  0-100          0.2200  Vogt 

(Olivin) 


140 


METALLURGICAL  CALCULATIONS. 


Substance 

Tem- 
perature 

Specific  Heat 

L.  H. 

Fusion 

L.  H. 

Vapor  i-     Authority 
zation 

Mg2SiO4 

1400° 

130 

Vogt 

(Olivin) 

Mg2SiO4  (liquid) 

1400 

[520 

Vogt 

toO°] 

MgSiO3 

0-t 

0.1974 

Vogt 

(Enstatite) 

+0.000086t 

MgSiO3 

1300 

125 

Vogt 

(Enstatite) 

MgSiO3  (liquid) 

1300 

[403 

Vogt 

toO°] 

CaSiO3 

0-t 

0.1690 

Vogt 

(Wollastonite) 

+0.0001t 

CaSiO3 

1250 



100 

Vogt 

(Wollastonite) 

CaSiO3  (liquid) 

1250 

[360 

.....  Vogt 

toO°] 

CaMgSi2O6 

0-t 

0.186 

Vogt 

+0.00008t 

CaMgSi2O6 

1225 

100 

Vogt 

CaMgSi2O6  (liquid) 

1225 

[344 

Vog 

toO°] 

Ca,MgSi4Ou 

0-t 

0.179 

Vog 

(Malacolite) 

+0.00007t 

Ca3MgSi4Oi2 

1200 

94 

Vogt 

Ca3MgSi4Oi2  (liquid 

)  1200 

[319 

Vogt 

toO°] 

H4Al2Si2O9 

20-98 

0.2243 

Ulrich 

(Kaolin) 

Al2Si(F)O6 

12-100 

0.1997 

Joly 

(Topaz) 

KAlSisOs 

20-100 

0.1877 

Oeber 

(Orthoclase) 

KAlSi3O8 

1200 

100 

Vogt 

(Orthoclase) 

KAlSi3O8 

20-100 

0.197 

Bogajaw- 

(  Micro  cline) 

lensky 

KAlSi3O8 

1170 

83 

Vogt 

(Microcline) 

CaAl2Si2O8 

0-t 

0.1790 

Vogt 

(Anorthite) 

+O.OQ01t 

CaAl2Si2O8 

1220 

100 

.....  Vogt 

(Anorthite) 

CaAl2Si2O8  (liquid) 

1220 

[358 

Vogt 

toO°] 

THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        141 


Substance 

Tent' 
perature 

Specific  Heat 

ZrSiO4 

15-100° 

0.1456 

(Zircon) 

BeAl2Si208 

12-100 

0.2066 

(Beryl) 

Fe3Al2Si3Oi2 

16-100 

0.1758 

(Iron  Garnet) 

Al2SiO6 

0-100 

0.1684 

(Audalusite) 

Asbestos 

20-98 

0.1947 

Serpentine 

16-98 

0.2586 

Talc  (soapstone) 

20-98 

0.2092 

Potash  mica 

20-98 

0.2080 

Sodium  mica 

20-98 

0.2085 

Magnesia  mica 

20-98 

0.2061 

Oligoclase 

20-98 

0.2048 

Spodumene 

20-100 

0.2176 

Labradorite 

20-98 

0.1949 

Hypersthene 

20-98 

0.1914 

Augite 

20-98 

0.1931 

Hornblende 

20-98 

0.1952 

Granite 

20-524 

0.2290 

Basalt 

0-t 

0.1937 

-f-0.000086t 

Basalt 

? 

Lava  (Etna) 

0-500 

0.1820 

+0.000155t 

Lava  (Etna) 

500-800 

0.270 

Pumice 

10-100 

0.240 

Ca,  K,  Glass 

14-99 

0.198 

Ca,  K,  Glass 

0-300 

0.190 

Flint  Glass 

10-50 

0.117 

Flint  Glass 

18-100 

0.082 

Crown  Glass 

10-50 

0.161 

Mirror  Glass 

10-50 

0.186 

Thermometer  glass 

0.1869 

(French,  hard) 

Jena  Glass 

18-99 

0.2182 

Thermometer  glass 

19-100 

0.1988 

(Normal,  German) 

Porcelain 

0-t 

0.2826 

-0.0000264t 

Cement  clinker 

28-40 

0.186 

Portland  cement 

28-30 

0.271 

Humus  (soil) 

20-98 

0.443 

Quartz  sand 

20-98 

0.1910 

L.  H. 

Fusion 


L.  H. 

Vapori-         Authority 
zation 

Regnault 


130 


Joly 

Oeberg 

Lindner 

Ulrich 

Oeberg 

Ulrich 

Ulrich 

Ulrich 

Ulrich 

Ulrich 

Schulz 

Ulrich 

Ulrich 

Ulrich 

Ulrich 

Bajrtoli 

Bartoli 

Tamman 
Bartoli 

Bartoli 
Dunn 
Regnault 
Dulong&  Petit 
H.  Meyer 
Winkelmann 
H.  Meyer 
H.  Meyer 
Zonboff 

Winkelmann 
Winkelmann 

Harker 

Hart! 

Haiti 
Ulrich 
Ulrich 


142 


METALLURGICAL  CALCULATIONS. 


Substance 

Sandstone 
Slag 

Enamel  slag 
Bessemer  slag 


Tem- 
perature 

0-100C 
14-99 
15-99 
14-99 


Specific  Heat 

0.174 
0.1888 
0.1865 
0.1691 


L.  H. 
Fusion 


L.  H. 

Vapor i-         Authority 
zation 


Hecht 
Oeberg 
Oeberg 
Oeberg 


MISCELLANEOUS  MATERIALS 


Paraffin  (solid)            0-t 

Paraffin  (liquid)        52-63 
Beeswax  (solid)         26-42 
Beeswax  (liquid)       65-100 
Soft  Para  Rubber       0-100 
Vulcanite                   20-100 
Anthracite                   0-12 
Oak  Wood                 

0.532 
+0rf0012t 
0.706 
0.82 
0.50 
0.48 
0.331 
0.312 
0.57 

Fir  Wood 

0.65 

Charcoal 

0  20 

Red  Brick                  

0.22 

Ether                             25 
Ether                             35 

0.54 

Alcohol                      20-26 
Alcohol  (liquid)            78 

0.58 

Benzol  (liquid)              10 
Benzol  (solid)                  5 

0.40 

Benzol  (liquid)             80 

Glycerin                     

0.58 

Glycerin                        13 

Machine  oil                

0.40 

Petroleum 

0  50 

Turpentine                
Turpentine                  159 

0.42 

Stearic  acid                  64 

Ammonia  (gas)           0-t 

Ammonia  (solution)     18 
Ammonia  (liquid)         16 

0.38 
+0.00016t 
1.00 

Methane  (CH4)          0-t 
Ethelyne  (C3H4)         0-t 

0.38 
+0.00022t 
0.46 
+0.0003t 

42.5 


47.6 


Battelli 

Battelli 
Person 
Person 
Gee  &  Terry 
A.  M.  Mayer 
Hecht 


, Regnault 

90    Brix 

Bose 

206     Schall 

Pickering 

30         J.  Meyer 

94    Tyrer 


Berthelot 


74    Brix 
.  .  .  .  Bruner 
.  LeChatelier 


Thomsen 

297     Regnault 


LeChatelier 


For  silicates  in  general,  the  specific  heat  is  found  to  agree 
fairly  well  with  what  would  be  calculated  from  their  percent- 


THERMOPHYSICS  OF  CHEMICAL  COMPOUNDS.        H3 

age  composition,  giving  each  oxide  constituent  its  proper 
specific  heat.  Those  oxides  whose  specific  heat  at  0°  are  known 
are  as  follows: 

SiO2     =  0.1833  (Richards) 

A1203  =  0.2081  (Richards). 

Fe2O3  =  0.1456  (Richards). 

MgO    =  0.2420  (Regnault). 

CaO     =  0.1779  (calculated). 

Those  which  are  not  known  give  good  results  if  assumed  to 
be  as  follows: 

FeO  =  0.1460  (Vogt). 
MnO  =  0.1511  (Vogt). 
K2O  =  0.1390  (Vogt). 
Na2O  =  0.2250  (Vogt). 
Li2O  =  0.4430  (Vogt). 

Using  these  data,  S  at  0°  is  calculated  from  the  percentage 
composition  of  the  silicate.  For  S  at  higher  temperatures  it 
can  be  assumed  with  considerable  approximation  to  the  truth, 
that  S  increases  0.078  per  cent,  for  each  degree,  and  so,  calling 
S0  the  specific  heat  at  zero,  we  would  have 

S  =  S0  (l+0.00078t) 

Sm  =  S0  (l+0.00039t) 

Q  (0°— 1°)  =  SmXt 

Besides  the  above  generalizations,  which  enable  one  to  cal- 
culate the  heat  in  the  solid  silicate  at  the  melting  point  (when 
the  latter  is  known),  Vogt  has  shown  that  if  the  heat  neces- 
sary to  heat  the  silicate  from — 273°  to  the  melting  point  is 
calculated,  the  latent  heat  of  fusion  may  be  taken  as  being  20 
to  25  per  cent,  of  this  quantity,  say  an  avejage  of  22.5  per  cent, 
and  thus  the  heat  required  for  fusion  may  be  approximately 
calculated. 

Akerman  has  determined  for  many  metallurgical  silicate 
slags  the  total  heat  contained  per  kilogram  of  melted  slag  at 
the  melting  point.  These  quantities  vary  from  347  to  530 
Calories,  depending  principally  on  the  elevation  of  the  melting 
point  of  the  slag.  A  brief  tabulation  of  Akerman 's  results  are 
as  follows,  arranged  according  to  the  amount  of  heat  in  the 
just-melted  slag: 


144 


METALLURGICAL  CALCULATIONS. 


Calories.                         % 
347..                               I 

SiO2 
59 

1 

350  

L39 

63 
58 
58 
53 

360  

41 
38 
39 
37 

66 
59 
48 
40 

380  

34 
31 
46 
58 

58 
62 
38 

• 
400  

25 

44 

60 
65 
41 

37 
21 
43 

%CaO 

%A1203 

36 

5 

42 

19 

35 

2 

35 

7 

37 

5 

37 

10 

42 

17 

47 

15 

43 

19 

40 

23 

32 

2 

38 

3 

42 

10 

48 

12 

48 

18 

37 

32 

37 

17 

32 

10 

27 

15 

37 

1 

52 

10 

34 

41 

33 

23 

20 

20 

35 

0 

52 

7 

53 

10 

32 

47 

30 

27 

The  above  includes  most  varieties  of  acid  and  basic  iron 
blast-furnace  slags.  .  Data  for  other  slags  made  in  the  metal- 
lurgy of  iron  and  of  other  metals,  are  almost  altogether  lack- 
ing. A  wide  field  is  here  open  for  metallurgical  experiments; 
data  thus  obtained  would  be  immediately  useful  in  practical 
calculations. 


CHAPTER  VI. 
ARTIFICIAL  FURNACE  GAS. 

There  are  many  different  forms  of  producers  for  making  artifi- 
cial furnace  gas.  For  the  purposes  of  making  calculations  upon 
them  they  may  be  conveniently  divided  into  four  classes,  as 
follows : 

1.  Simple  producers,  those  which  use  ordinary  fuels,  such  as 
wood,  peat,  lignite,  bituminous  coal  or  anthracite,  and  in  which 
no  water  or  water  vapor  is  introduced  other  than  the  water  in 
the  fuel  itself  and  the  normal  moisture  of  the  air  used. 

2.  Mixed  gas  producers,  in  which  water  vapor  or  steam  is 
introduced  with  the  air  for  combustion,  in  such  amount  as  to 
be  entirely  decomposed  in  passing  through  the  fuel. 

3.  Mond  gas  producers,  in  which,  for  a  special  purpose,  more 
steam  is  introduced  than  can  be  decomposed  in  the  producer, 
thus  producing  very  wet  gas. 

4.  Water  gas  producers,  in  which  air  and  steam  alone  are 
alternately  fed  to  the  producer,  the  former  for  heating  up,  the 
latter  for  producing  water  gas. 

As  far  as  the  calculations  are  concerned,  the  essential  dif- 
ference between  these  classes  is  the  varying  amount  of  water 
vapor  or  steam  introduced  under  the  fuel  bed  while  producing 
the  gas,  from  all  air  in  Class  1  to  all  steam  in  Class  4. 

The  calculations  which  it  is  of  immediate  interest  to  make,  and 
the  results  of  which  are  of  immediate  value  to  the  metallurgist, 
are  those  concerned  with  the  volume  of  gas  produced  per  unit  of 
fuel,  its  calorific  power  compared  to  that  of  the  fuel  from  which 
it  is  produced,  the  items  of  the  heat  losses  during  the  operation 
of  transforming  the  solid  fuel  into  gaseous  fuel,  the  function 
of  steam  in  the  producer,  the  limits  up  to  which  the  use  of  steam 
is  permissible,  the  increase  of  efficiency  of  the  gas  by  subsequently 
drying  it,  the  advantages  as  to  final  efficiency  which  are  gained 
by  gasifying  the  fuel  over  burning  solid  fuel  directly. 

145 


146  METALLURGICAL  CALCULATIONS. 

For  information  as  to  the  construction  and  operation  of  gas 
producers,  reference  may  be  made  to  treatises  such  as  Sexton's 
or  Wyer's  "  Producer  Gas,"  Groves  and  Thorp's  "  Chem- 
ical Technology,  Vol.  V.,  Fuels."  Trade  pamphlets  and  cata- 
logues, such  as  those  of  R.  D.  Wood  and  Co.  on  "  Gas  Pro- 
ducers," of  the  De  la  Vergne  Machine  Co.  on  "  Koerting  Gas 
Engines  and  Gas  Producers,"  etc.,  contain  a  great  deal  of 
exact  and  useful  information,  and  may  be  usually  had  for  the 
asking.  The  monograph  of  Jiiptner  and  Toldt  on  "  Genera - 
toren  und  Martinofen  "  (Felix,  Leipzig,  1900),  is  concerned 
wholly  with  calorimetric  calculations  concerning  the  produc- 
tion of  gas  and  its  utilization  in  regenerative  gas  furnaces. 

1. — SIMPLE  PRODUCERS. 

In  these  a  deep  bed  of  fuel  is  burnt  by  air  or  fan  blast,  in- 
troducing no  more  moisture  than  happens  to  be  in  the  atmo- 
sphere at  the  time  being.  The  fuel  fed  into  the  producer  is  first 
dried  by  the  hot  gases,  then  is  heated  and  distilled  or  coked, 
and  finally  is  oxidized  by  the  incoming  air.  The  residue  is 
the  ash  of  the  coal,  which  is  ground  out  at  the  bottom  or  drops 
through  the  grate,  containing  more  or  less  unburnt  fixed  car- 
bon. Great  loss  of  efficiency  sometimes  occurs  from  the  ashes 
being  rich  in  carbon.  The  escaping  gases  issue  at  tempera- 
tures of  300°  up  to  1000°  C.,  carrying  much  sensible  heat  out 
of  the  producer. 

Calculations  as  to  the  amount  of  gas  produced  per  unit  of 
fuel  consumed  are  to  be  based  entirely  on  the  carbon.  The  gas 
mtist  be  carefully  analyzed,  so  that  it  can  be  calculated  from 
this  analysis  how  much  carbon,  in  weight,  is  contained  in  a 
given  volume  of  gas.  (An  alternative  method  is  to  take  a 
carefully  measured  volume  of  the  gas,  mix  it  with  excess  of 
oxygen,  and  explode  in  a  gas  burette,  determining  the  amount 
of  carbon  dioxide  formed,  and  from  that  calculate  the  weight 
of  carbon  in  the  volume  of  gas  taken.)  Knowing  this,  the 
rest  is  simple;  the  carbon  in  unit  weight  of  fuel  minus  the 
carbon  lost  in  the  ashes  by  poor  combustion  gives  the  weight 
of  carbon  gasified ;  this  divided  by  the  weight  of  carbon  in 
unit  volume  of  gas  produced,  gives  the  volume  of  the  latter 
per  unit  weight  of  fuel. 

Illustration',    A  fuel  used  in  a  gas  producer  contains  12  per 


ARTIFICIAL  FURNACE  GAS.  147 

cent,  of  ask  and  72  per  cent,  of  carbon.  The  ashes  made  con- 
tain 20  per  cent,  of  unburnt  carbon;  the  gas  produced  contains 
by  analysis  and  calculation  0.162  ounce  of  carbon  per  cubic 
foot  of  gas,  measured  at  60°  F.  and  29.8  inches  barometric 
pressure.  What  volume  of  gas,  measured  at  above  conditions, 
is  being  produced  per  ton  of  2240  pounds  of  coal  used? 

Solution:  The  distinction  between  ash  and  ashes  must  be 
noted;  the  former  is  the  analytical  expression  for  the  amount 
of  inorganic  material  left  after  complete  combustion  during  the 
chemical  analysis;  the  latter  term  means  the  waste  matter  pro- 
duced industrially,  and  consists,  if  weighed  dry,  of  the  true 
ash,  plus  any  unburnt  carbon.  The  calculations  are  therefore 
as  follows :  T  . 

Ash  in  2240  pounds  of  coal  =  2240  X  0. 12  =  268 . 8 

Ashes  corresponding  =  268.8-^0.80  =  336.0 

(the  ashes  are  80  per  cent,  ash) 

Carbon  in  the  ashes  =  67 . 2 

Carbon  in  the  coal  =  2240X0.72  =  1612.8 

Carbon  going  into  the  gas  (gasified)  =  1545.6 

Carbon  in  1  cubic  foot  of  gas  =  0.162-7-16  0.010125 
Volume  of  gas  produced  per  2240  pounds  of 

coal  =  1545.6 -=-0.010125  =  152,652  cu.  ft 

It  will  be  noted  that  these  calculations  absolutely  require 
the  percentage  of  total  carbon  in  the  fuel,  as  determined  by 
chemical  analysis.  This  is  not  a  difficult  analysis,  as  it  consists 
in  burning  the  carbon  in  a  heated  tube  in  a  stream  of  oxygen 
or  air  free  from  carbon  dioxide;  the  products  of  combustion 
are  dried  and  then  passed  through  caustic  potash  solution  to 
absorb  CO2  gas,  the  weight  of  which  is  obtained  by  the  in- 
creased weight  of  the  potash  bulb,  and  the  total  carbon  thus 
obtained.  The  fixed  carbon  and  volatile  matter  of  the  coal,  as 
determined  by  the  ordinary  proximate  analysis,  cannot  be  used 
in  this  calculation,  since  while  all  the  fixed  carbon  is  carbon, 
the  volatile  matter  is  of  variable  composition,  containing  such 
varying  proportions  of  carbon  that  no  fixed  percentage  of  the 
latter  in  it  can  be  assumed  without  considerable  possible  error. 

The  calorific  power  of  the  gas,  per  cubic  meter  or  cubic 
foot,  can  be  calculated  from  its  analysis,  using  the  calorific 
powers  of  the  combustible  constituents  as  already  given  in  our 


148  METALLURGICAL  CALCULATIONS. 

tables.  This,  multiplied  by  the  volume  of  gas  produced  per 
unit  of  coal,  gives  the  calorific  power  of  the  gas  as  compared 
with  that  of  the  coal  from  which  it  is  made.  The  difference 
is  the  heat  loss  in  the  operation  of  producing  the  gas,  including 
loss  by  unburnt  carbon  in  the  ashes.  In  fact,  we  may  state  that 
the  heat  balance  is  based  on  the  following  equations: 

Heating  power  of  the  coal,  per  unit 
— Heating  power  of  the  gas  per  unit  of  coal 
=  Calorific  losses  in  conversion. 

The  latter  item  is  composed  of: 

Loss  by  unburnt  carbon  in  the  ashes. 
Sensible  heat  of  the  hot  gases  issuing. 
Heat  conducted  to  the  ground. 
Heat  radiated  to  the  air. 

These  items  may  be  modified  as  follows:  If  the  air  used  is 
hotter  than  the  normal  outside  temperature,  its  sensible  heat 
above  this  datum  should  be  added  to  the  heating  power  of  the 
coal,  because  it  increases  the  total  available  heat.  If  the  ashes 
are  removed  hot,  and  not  allowed  to  be  completely  cooled  by 
the  incoming  air,  their  sensible  heat  should  be  included  in  the 
calorific  losses  during  conversion.  If  the  air  used  is  moist,  its 
moisture  will  be  decomposed  to  hydrogen  and  oxygen,  but  the 
heat  absorbed  in  doing  this  is  exactly  represented  by  the 
calorific  power  of  this  increased  amount  of  hydrogen  in  the 
gases,  and  the  heat  absorbed  is  not  lost  but  really  represents  so 
much  saved  as  available  calorific  power  of  the  gases.  This  item 
must,  therefore,  not  be  counted  as  one  of  the  heat  losses  during 
the  operation,  as  those  losses  have  been  defined  by  us.  If  the 
fuel  is  wet,  considerable  heat  is  required  to  evaporate  the  mois- 
ture in  it,  but  this  heat  is  not  to  be  reckoned  as  one  of  the 
losses  in  conversion,  if  we  have  taken  as  the  heating  power  of 
the  coal  the  practical  metallurgical  value;  that  is,  its  value 
assuming  all  the  water  in  its  products  of  combustion  to  re- 
main as  vapor  and  none  to  condense.  If  this  value  has  been 
so  taken,  the  heat  required  to  vaporize  the  moisture  in  the 
coal  will  have  already  been  allowed  for.  Similarly,  it  may  take 
a  little  heat  energy  to  break  up  a  bituminous  coal  so  as  to 
expel  its  volatile  matter,  but  this  should  not  be  reckoned  in  as 


ARTIFICIAL  FURNACE  GAS.  149 

a  heat  loss  in  the  producer,  because  a  little  reflection  will  show 
that,  whatever  this  amount  may  be,  it  has  been  properly  al- 
lowed for  in  the  determination  or  calculation  of  the  total  calo- 
rific power  of  the  fuel.  Jiiptner  and  Toldt  call  this  the  "  gas- 
ifying heat,"  and  use  it  in  all  their  calculations,  but  it  is  doubt- 
ful whether  it  really  amounts  to  an  appreciable  quantity,  for 
one  thing,  and  even  if  it  does  it  should  not  be  reckoned  as  a 
heat  loss  in  the  producer. 

Problem  16. 

Jiiptner  and  Toldt  ran  a  gas  producer  with  lignite  of  the 
following  composition:  (Generatoren,  p.  49). 

Carbon 69.83  per  cent. 

Hydrogen 4.33 

Nitrogen 0.50 

Oxygen 12.38 

Moisture 7.25 

Ash 5.71 

Of  this  coal,  3214  kilograms  was  used  in  8  hours,  50  minutes, 
producing  gas  which  contained,  analyzed  dry,  by  volume- 

Carbon  dioxide,  CO2. 5. 21  per  cent. 

Carbon  monoxide,  CO 23.99 

Oxygen,  O2 0.63 

Methane,  CH4 0.25 

Hydrogen,  H2 10.64 

Nitrogen,  N2 59.28 

The  ashes  produced  weighed  22.23  kilograms  per  100  kilo- 
grams of  coal  used,  and  contained  68.76  per  cent,  of  unburned 
carbon.  The  calorific  power  of  the  coal,  determined  in  the 
calorimetric  bomb  (in  compressed  oxygen,  moisture  resulting 
condensed)  was  6949  Calories  per  kilogr.  Temperature  of 
hot  gases,  282°  C.;  temperature  of  air  used,  9°  C.;  humidity, 
62  per  cent.;  barometer,  712  millimeters  of  mercury. 
Required : 

1.  The  volume  of  gas,  measured  at  0°  and  760  mm.  pressure 
(and  assumed  dry),  produced  per  metric  ton  (1000  kilos.  = 
2204  pounds)  of  fuel  used. 


150  METALLURGICAL  CALCULATIONS. 

2.  The  calorific  power  of  the  coal,  per  kilogram,  with  mois- 
ture formed  by  its  combustion  assumed  uncondensed. 

3.  The  proportion  of  the  calorific  power  of  the  coal  develop- 
able by  burning  the  gas  produced  from  it. 

4.  The  loss  of  heat  in  conversion. 

5.  The  loss  of  heat  by  unburnt  carbon  in  the  ashes. 

6.  The  loss  of  heat  as  sensible  heat  in  the  gases. 

7.  The  loss  of  heat  by  radiation  and  conduction,  expressed: 

(a)  Per  unit  of  coal  burnt. 

(b)  Per  minute. 

8.  The  volume  of  air  required  by  the  producer,  at  the  con- 
ditions of  the  atmosphere,  per  kilogram  of  coal  burnt. 

Solution  : 

(1)  The  gas  contains,  per  cubic  meter  at  standard  condi- 
tions : 

Carbon  in  CO2  0.0521  X  0.54  kilos. 
Carbon  in  CO   0.2399x0.54     " 
Carbon  in  CH4  0.0025  X  0.54     " 

Total  0.2945X0.54  =  0.1590  kilos. 
The  carbon  gasified  from  1000  kilograms  of  coal  is: 

Carbon  in  coal  =  698.3  kilos. 

Carbon  in  ashes  222.3X0.6876  =  152.8     " 

Carbon  gasified  =  545.5     " 

Therefore, 

C/1K      C 

Gas  (dry)  produced  =  ^     "^    =  3430  cubic  meters.     (1) 


(2)  The  calorific  power  of  the  coal  as  given  must  be  dimin- 
ished by  the  heat  required  to  vaporize  all  the  moisture  formed 
by  its  combustion,  leaving  such  moisture  as  theoretical  mois- 
ture at  0°  C.  There  will  be  formed  per  kilogram  of  coal: 

From  moisture  of  coal  =  0.0725  kilos. 

From  hydrogen  0.0433x9  =  0.3897     " 

Total  =  0.4622     " 
To    evaporate    this   to    theoretical   moisture    at    zero    (thus 


ARTIFICIAL  FURNACE  GAS.  151 

putting  the  water  vapor  on  the  same  footing  as  the  other  pro- 
ducts of  combustion,  CO2  and  N2)  requires: 

0.4622X606.5  (Regnatdt),  =  280  Calories, 
leaving  as  the  metallurgical  or  practical  calorific  power 

6949—280  =  6669  Calories,  (2) 

(3)  The  calorific  power  of  each  cubic  meter  of  gas   (meas- 
ured dry  at  standard  conditions)  is 

CO    =  0.  2399  m3  X  3,062  =  734  .  6  Calories 

CH4  =  0.0025  m3X  8,598  =  21.5 

H2     =  0.1064  m3X  2,613  =  278.0 

Total  =  1034.1 

Calorific  power  of  gas  from  1  kilogram  of  coal: 
1034.1X3.43  =  3547.0  Calories. 

which  equals      F-    =  53.2  per  cent.  (3) 


(4)  The  loss  of  calorific  power  in  conversion  is  100  —  53.2 
=  46.8  per  cent,  of  the  calorific  power  of  the  coal,  or  per  kilo- 
gram of  coal: 

6669—3547  =  3122  Calories.  (4) 

(5) 
Carbon  in  ashes  =  0.1528X8100  =  1237.7  Cal. 

=       18.6  per  cent.  (5) 

(6)  The  gases  produced  carry  off,  per  cubic  meter  measured 
dry,  the  following  amounts  of  heat: 

Volume  X  mean  specific  heat  (0°  —  282°)  =  heat  capacity 
per  1°. 

CO2  0.0521X0.432  =  0.0225 
CH4  0.0025X0.428  =  0.0011 
CO 


0.9454X0.311  =  0.2940 


N2  Sum  =  0.3176 

Heat  carried  out  =  0.3176X282  =  89.56  Calories. 
Per  kilogram  of  coal  =  89.56x3.43  =  307  Calories. 

307 
Proportion  of  calorific  power  =  ^^  =  4.6  per  cent. 


152  METALLURGICAL  CALCULATIONS. 

The  above  result  is,  however,  subject  to  a  small  correction, 
because  some  of  the  moisture  in  the  coal  goes  undecomposed 
into  the  gases,  and  is  not  represented  in  the  analysis  of  the 
dried  gas.  The  amount  of  this  moisture  can  be  obtained  with 
sufficient  accuracy  by  finding  how  much  moisture  would  be 
obtained  by  burning  the  dried  gas  from  1  kilogram  of  coal, 
and  comparing  this  with  the  moisture  which  would  be  ob- 
tained from  1  kilogram  of  coal  itself;  the  difference  must  rep- 
resent the  moisture  accompanying  the  gas  as  water  vapor, 
and  which  has  not  been  included  in  the  above  computation. 

Burning  1  cubic  meter  of  gas,  the  H20  vapor  is: 

From  CH4  0.0025X2  =  0.0050  cubic  meters. 

FromH2    0.1064X1  =  0.1064     " 

0.1114     " 
Per  kilo,  of  coal  =  0.1114X3.43  =  0.3821      " 

But  the  weight  of  water  vapor  from  burning  1  kilogram 
of  coal  has  already  been  found  to  be  (2)  0.4622  kilograms,  the 
volume  of  which  is 

0.4622 -r- 0.81  =  0.5706  cubic  meters. 

Leaving,  therefore,  0.5706—0.3821  =  0.1885  cubic  meters  of 
water  vapor  as  such  accompanying  the  3.43  cubic  meters  of 
(dried)  gas  from  1  kilogram  of  coal.  This  would  take  out 

0.1885X0.382X282  =  20.3  Calories. 
Thus  increasing  the  sensible  heat  in  the  gases  to 

307  +  20.3  =  327.3  Calories  =  4.9  per  cent.  (6) 

[In  reality,  a  still  further  correction  should  be  made;  viz.: 
to  add  in  the  moisture  in  the  air  used,  because  it  would  also 
reappear  as  moisture  on  final  combustion  of  the  gases.  Its 
amount  is  found  from  the  amount  of  air  used,  which,  if  62  per 
cent,  saturated  at  9°  would  carry  moisture  having  0.62  X 
8.6  mm.  (if  saturated)  =  5.3  millimeters  tension,  which  repre- 
sents, barometer  being  712  mm.,  0.7  per  cent,  of  the  volume 
of  the  air  used,  or  practically  0.9  per  cent,  of  the  volume  of 
nitrogen  in  the  air.  Since  the  nitrogen  in  the  gas  represents 
almost  entirely  the  nitrogen  in  the  air  used,  the  moisture  to  be 
accounted  for  from  the  air  amounts  to  0.5928X0.009  =  0.0053 


ARTIFICIAL  FURNACE  GAS.  153 

cubic  meters  per  cubic  meter  of  gas,  or  =  0.0182  cubic  meters 
per  kilogram  of  coal  burnt.  This  correction  is  altogether  too 
small  to  affect  the  results  in  this  case,  but  should  be  taken 
into  account  whenever  the  air  used  is  warm  and  moist.] 

(7)  The  calorific  loss  in  conversion  was  3122  Calories.  Of 
this  we  have  accounted  for: 

Lost  by  unburnt  carbon  in  ashes 1237.7  Calories. 

Sensible  heat  of  gases-  (including  moisture) 327.3 

Total 1565.0 

Loss  by  radiation  and  conduction =  1557.0         "  (a) 

Per  8  hours  5  minutes  there  is  burnt  3214  kilograms  of 
fuel,  making  the  loss  of  heat  by  radiation  and  conduction  per 
minute  = 

1557X3214 


530 


=  9,442  Calories.  (b) 


(8)  At  the  conditions  given,  each  cubic  meter  of  moist  air 
used  contained  0.7  per  cent,  of  its  volume  of  moisture,  making 
its  percentage  composition  by  volume: 

Water  vapor 0.70  per  cent. 

j  Oxygen 20.65 

Air  (  Nitrogen 78.65 

The  volume  of  gas  produced  per  kilogram  of  coal  is  3.43  cubic 
meters,  of  which  59.28  per  cent,  is  nitrogen,  equal  to  2.0333 
cubic  meters,  and  weighing  2.0333x1.26  =  2.562  kilograms. 
Of  this  0.0050  kilograms  came  from  the  coal  itself,  leaving 
2.512  kilograms  to  come  from  the  air,  or  2.512 -i- 1.26  =  1.9921 
cubic  meters.  This  would  correspond  to  1.9921 -=-0.7865  = 
2.5329  cubic  meters  of  moist  air  if  measured  at  standard  con- 
ditions. At  9°  and  712  mm.  pressure  the  real  volume  of  moist 
air  used  at  prevailing  conditions,  per  kilogram  of  coal  burnt,  is 

7fif) 

^  =  2.80  cubic  meters.       (8) 

It  must  not  be  thought  that  the  conditions  of  working  in 
the  above  producer  represent  good  practice;  they  are  very 
poor  practice  as  far  as  concerns  the  utilization  of  the  fuel. 
Many  producers  make  gas  having  75  to  90  per  cent,  of  the 


154  METALLURGICAL  CALCULATIONS. 

calorific  power  of  the  coal  from  which  it  is  made,  so  that  the 
losses  by  unburnt  carbon  and  radiation  and  conduction  in  this 
case  must  be  regarded  as  highly  abnormal  and  very  poor  prac- 
tice. The  writer  chose  this  example  for  calculating,  because  of 
the  carefulness  with  which  Juptner  and  Toldt  had  collected  the 
necessary  data,  and  because  it  illustrated  so  well  the  principles 
to  be  employed  in  similar  calculations. 

Problem  17. 

A  gas  producer  run  in  Sweden  uses  saw-dust  of  the  following 
composition : 

Water 27.0  per  cent. 

Ash 0.5 

Carbon 37.0 

Hydrogen 4.4 

Oxygen . 30.6 

Nitrogen 0.5 

Assume  that  it  is  run  by  dry  air  and  that  0.5  per  cent,  of 
ashes  are  made.  The  gas  formed,  dried  before  analysis,  con- 
tains, by  volume: 

Carbon  dioxide,  CO2 ,  .   6.0  per  cent. 

Carbon  monoxide,  CO 29.8 

Ethylene,  C2H4 0.3 

Methane,  CH4 6.9 

Hydrogen,  H2 6.5 

Nitrogen,  N2 50.5 

The  gas  actually  produced  is  -partly  dried  before  use  by 
having  its  temperature  reduced  by  cold  water,  in  a  surface 
condenser,  to  29°  C.,  in  order  to  increase  its  calorific  intensity 
of  combustion. 

Required: 

(1)  The  proportion  of  the  moisture  in  the  moist  gas  which 
is  condensed  out. 

(2)  The  calorific  intensity  of  the  moist  gas,  if  burned  pre- 
heated to  800°  C.  by  the  theoretical  quantity  of  air  preheated 
also  to  800°  C. 


ARTIFICIAL  FURNACE  GAS.  155 

(3)  The   calorific  intensity  of  the   dried  gas,   burnt   under 

exactly  similar  conditions. 

Solution : 

(1)  It  is  first  necessary  to  find  the  weight  or  volume  of 
water  vapor  accompanying  the  gas  before  condensation,  next 
that  accompanying  it  after  passing  the  condenser.  The  first 
can  be  calculated  best  on  the  basis  of  the  hydrogen  present 
in  the  fuel  and  in  the  (dried)  gas  made  from  it;  the  difference 
is  the  hydrogen  of  the  moisture  removed  before  analysis,  i.e., 
the  hydrogen  of  the  moisture  in  the  wet  gas. 

The  first  step  is  to  find  the  volume  of  gas  (dry)  produced 
per  unit  of  fuel,  as  follows: 

Carbon  in  1  kilo,  of  fuel  =  0.370    kilos. 

Carbon  in  1  m3  of  gas  =  CO2  +  CO  +  CH4  + 
2C2H4  (0.060 +  0.298 +  0.069  + 0.006)  X  0.54     =  0.2338     " 

Dry  gas  per  kilo,  of  fuel  =  •     '       -  =  1.5825  m3 

U . Zooo 

The  next  step  is  to  calculate  the  water  which  would  be  formed 
by  the  combustion  of  1  kilogram  of  fuel: 

Water  present  in  fuel 0. 270  kilos. 

Water  produced  by  hydrogen Q.  396     " 

Total  =  0.666     " 

Volume  at  standard  conditions  =     '         =  0.8222  me 

0 .  o  1  ( ) 

From  this  we  subtract  the  moisture  which  would  be  pro- 
duced by  the  combustion  of  the  1.5825  cubic  meters  of  dry  gas, 
obtained  as  follows: 

Water  from  ethylene    =  0.003  X  2  X  1.5825  m3 
Water  from  methane    =  0.069 X2X  1.5825 
Water  from  hydrogen  =  0.065X1  X  1.5825 

=  0.209       XI. 5825  =  0.3307m8 
Difference  =  water  vapor  to  1.5825  m3  of  dried  gas 
=  0.8222—0.3307  =  0.4915m3 
=  0.4915-5-1.5825  =  0.3106  m3  per  1  m8 

of  dried  gas,  as  analyzed 


156  METALLURGICAL  CALCULATIONS. 

This  is  to  be  compared  with  the  amount  of  moisture  ac- 
companying the  same  quantity  of  (dried)  gas  as  it  escapes 
from  the  condenser.  This  is  obtained  directly  from  the  fact 
that  the  gas  escaping  will  be  saturated  with  moisture  at  29° 
C.,  that  the  latter  will,  therefore,  have  a  tension  of  30  milli- 
meters (tables),  and  that,  assuming  the  barometer  normal 
(760  mm.),  the  partial  tensions  of  moisture  and  gas  proper 
are  as  30  to  760—30,  or  as  30  to  730.  Since  their  respective 
volumes  (if  both  were  measured  separately  at  normal  pres- 
sures) are  in  the  same  proportion,  it  follows  that  each  cubic 
meter  of  (dry)  gas  is  accompanied  by  30  -r-  730  =  0.0411  cubic 
meters  of  uncondensed  moisture. 

The  respective  quantities  of  moisture  accompanying  1  cubic 
meter  of  dry,  uncondensable  gas,  are,  therefore,  0.3106  before 
cooling  and  0.0411  after  cooling,  showing  that  13.2  per  cent,  of 
all  the  moisture  escapes  condensation,  and  that,  therefore, 

86.8  per  cent,  of  moisture  is  condensed.  (1) 

(2)  The  wet  gas  has  a  calorific  power,  calculating  on  the 
basis  of  1  cubic  meter  of  dried  gas  analyzed: 

CO     0.298 X   3,062  =  912.5  Calories. 

C2H4  0.003  X  14,480  =  43.4 

CH4  0.069  X   8,598  =  593.3 

H2     0.065  X   2,613  =  169.8 

Total  =  1719.0 

There  is  added  to  this  available  heat,  when  burned,  the 
sensible  heat  in  the  gas  itself,  at  800°  C.,  and  also  that  of  the 
necessary  air,  also  at  800°. 

Heat  in  gas: 

CO,H2,N2  =  0.868    m3X 0.3246  =  0.2817  Cals.  per  1° 

CO2  =  0.060    m3X  0.5460  =  0.0328 

CH4  =  0.069    m3X  0.4485  =  0.0309 

C2H4  =  0.003    m3X0.50      =  0.0015 

H20  =  0.3106  m3X  0.460    =0.1429 

Calorific  Capacity  =  0.4898 
Total  sensible  heat  =  0.4898X800  =  391.8  Calories. 


ARTIFICIAL  FURNACE  GAS.  157 

The  air  required  theoretically  is: 

For  CO     0.298  m3  =  0.1490  m3  oxygen 
For  C2H4  0.003  m3  =  0.0090  m3 
For  CH4  0.069  m3  =  0.1380  m3 
For  H2      0.065  m3  =  0.0325  m3 
Sum  =  0.3285  m3 
=  1.58m3  air 

Heat  in  this  at  800°  =  1.58X0.3246X800  »  410.3  Calories. 
Sum  total  of  heat  going  into  the  products: 

Developed  by  combustion 1719.0  Cals. 

Sensible  heat  in  gas 391.8     " 

Sensible  heat  in  air 410.3     " 

Total 2520.0     " 

The  products  of  the  combustion  are  CO2,  H2O  and  N2,  as 
follows:  CO2  =  0.060  (in  gas) +0.298  (from  CO) +0.006  (from 
C2H4)+ 0.069  (from  CH4)  =  0.433m3. 

H20  =  0.311  (with  gas) +0.006  (from  C2H4) +0.138  (from 
CH4)+ 0.065  (from  H2)  =  0.520m3. 

N2  =  0.505  (in  gas) +[1.58— 0.33  =  1.25]  (from  air)  =  1.755 
m3. 

Since  the  2520.0  Calories  remains  as  sensible  heat  in  the 
above  products,  at  some  temperature  t,  we  have 

Heat  capacity  of  the  CO2  =  0.433  (0.37  +0.00022t) 

Heat  capacity  of  the  H2O  =  0.520  (0.34  +0.00015t) 

Heat  capacity  of  the  N2  =  1.755  (0.303  +  0.000027t) 

Heat  capacity  of  the  products     =      0.8688  +  0.00022065t 
and  the  calorific  intensity  t  must  be 

2520.0 

t  "  0.8688  +  0.00022065t 
whence 

t  =  1942°.  (2) 

(3)  When  the  dried  gas  is  burned  under  similar  conditions, 
the  only  difference  is  that  0.2695  m3  of  water  vapor  are  absent 
from  the  gas  and  from  the  products,  having  been  condensed. 


158  METALLURGICAL  CALCULATIONS. 

This  reduces  the  available  heat  by  the  sensible  heat  in  this 
much  water  vapor  at  800°,  viz. : 

0.2695X0.460X800  =  99.2  Calories, 
and  decreases  the  calorific  capacity  of  the  products  by 

0.2695  (0.34  +  0.00015t). 
Our  equation,  therefore,  becomes 

=  2360.2 

"  0.7772  +  0.00018023t 

whence 

t  -  2056° 

The  increased  efficiency  of  the  dried  gas  for  obtaining  high 
temperatures  is  too  evident  to  need  further  comment,  the 
difference  being  in  round  numbers  150°  C.,  equal  to  270°  F.f 
in  favor  of  the  dried  gas. 


2.  MIXED  GAS  PRODUCERS. 

This  Ciass  of  producers  are  those  most  commonly  used.  -In 
them  a  moderate  amount  of  steam  or  vapor  of  water  passes 
with  the  air  into  the  fire,  and  is  decomposed,  producing  carbon 
monoxide  and  hydrogen  gases  by  the  reaction: 

i  i         i 

H2O  +  C  =  CO  +  H2 

18       12       28        2 

which  may  be  read  as  follows:  One  volume  of  steam  forms 
one  volume  of  carbon  monoxide  and  one  volume  of  hydrogen; 
or  eighteen  parts  by  weight  of  water  vapor  act  upon  twelve 
parts  of  solid  carbon,  producing  twenty-eight  parts  of  carbon 
monoxide  and  two  parts  of  hydrogen.  If  we  speak  of  kilo- 
grams as  the  above  weights,  then  we  can  call  each  "  volume  " 
spoken  of  22.22  cubic  meters:  or  if  we  call  the  weights  ounces 
avoirdupois,  each  "  volume  "  represents  22.22  cubic  feet. 

The  water  vapor  is  admitted  either  automatically,  as  in  the 
old  Siemen's  type  of  producer,  where  water  was  run  into  the 
ash  pit  to  be  evaporated  by  the  heat  radiated  from  the  grate  or 
by  hot  ashes  falling  into  it,  or  as  in  the  modern  water-seal 


ARTIFICIAL  FURNACE  GAS.  159 

bottom  producer,  where  the  ashes  rest  upon  water  in  a  large 
pan,  and  so  are  continually  kept  soaked  by  capillary  action; 
or,  finally,  steam  is  positively  blown  under  the  grate,  either 
as  a  simple  steam  jet,  or,  more  economically,  by  using  it  in  a 
steam  blower,  so  as  to  have  it  produce  by  injector  action  an  air 
blast  sufficient  to  run  the  producer.  In  the  latter  case  the 
proportions  of  air  and  steam  may  be  regulated  with  precision, 
and  the  blast  action  produced  makes  the  production  and  de- 
livery of  the  gas  practically  independent  of  chimney  dralt.  The 
use  of  steam  also  rots  or  disintegrates  the  ashes,  preventing 
or  breaking  up  masses  of  clinker,  and  so  facilitating  the  re- 
moval of  the  ashes. 

Steam  or  water  vapor  cools  down  the  fire  in  the  producer 
so  that  it  runs  cooler;  at  the  same  time  gas  is  produced  which 
is  rich  in  hydrogen,  and,  therefore,  of  higher  calorific  power. 
This  saves  unnecessary  waste  of  heat  in  the  producer,  and  in- 
creases the  efficiency  of  the  gas  in  the  furnace  in  which  it  is 
burnt.  The  scientific  reasons  for  these  facts  are  to  be  found 
in  a  consideration  of  the  thermochemistry  of  the  reaction  by 
which  steam  is  decomposed. 

H2O  +  C  =  CO  +  H2 
—69,000    +29,160    = —39,840  Calories. 

This  would  be  the  deficit  in  decomposing  18  kilograms  of 
water  if  it  starts  in  the  liquid  state.  If,  however,  it  is  used  as 
steam  at  100°  C.,  each  kilogram  contains  637  Calories  of  sensi- 
ble heat,  making  18X637  =  11,466  Calories  altogether,  leav- 
ing the  deficit  28,374  Calories,  or  the  deficit 

=  1,576  Calories  per  kilogram  of  steam  decomposed. 
=  2,364  Calories  per  kilogram  of  carbon  thus  burnt. 

In  making  this  calculation  it  might  be  objected  that  the 
steam  used  is  often  at  three  or  four  atmospheres  pressure,  and 
its  temperature,  therefore,  over  100°  C. ;  but  it  must  not  be 
overlooked  that  this  steam  expands  suddenly  to  atmospheric 
tension,  and  that  in  so  doing  it  cools  itself  to  an  amount  roughly 
proportional  to  its  excess  pressure  so  that  the  expanded  steam 
at  atmospheric  pressure  is  usually  close  to  100°  C.  When 
mixed  with  air,  the  temperature  of  the  mixture  is  usually 
below  100°,  some  40°  to  50°  C.,  and  in  this  condition  some  of 


160  METALLURGICAL  CALCULATIONS. 

the  steam  is  possibly  condensed  to  fog,  but  it  must  not  be  for- 
gotten that  the  heat  of  condensation  thus  given  out  has  been 
absorbed  in  raising  the  temperature  of  the  admixed  air,  and 
therefore  goes  as  sensible  heat  into  the  bed  of  burning  fuel. 
For  the  purpose  of  calculation,  we  will,  therefore,  be  very 
nearly  right  in  assuming  that  we  are  dealing  in  each  case  with 
steam  at  100°  C.,  requiring  the  above  calculated  deficits  to  be 
made  up,  in  order  for  the  decomposition  to  proceed. 

It  will  be  evident  that  any  heat  thus  used  in  the  producer 
must  be  sensible  heat  of  the  hot  carbon,  which,  if  not  so  used, 
would  be  lost  as  sensible  heat  in  the  gases  produced  or  radi- 
ated and  conducted  away  from  the  producer.  The  basic  pro- 
cess of  the  running  of  the  producer  is  the  burning  of  carbon 
ttf  carbon  monoxide,  liberating  2,430  Calories  per  kilogram  of 
carbon,  which  is  2,4304-8,100  =  30  per  cent,  exactly  of  the 
calorific  power  of  the  carbon.  This  heat,  if  no  water  vapor 
gets  into  the  producer,  is  lost  as  sensible  heat  of  the  hot  gases 
and  by  radiation  and  conduction,  and  is  thus  largely  a  dead 
loss.  In  Problem  16,  for  instance,  these  items  figured  out  28.25 
per  cent,  of  the  calorific  power  of  the  coal  used.  Now,  the 
facts  are,  that  while  small  producers  need  that  much  heat  to 
keep  them  up  to  working  temperature,  large  producers  need 
very  much  less,  and  run  far  too  hot  if  no  steam  is  admitted 
to  check  the  rise  of  temperature.  In  the  largest, producers  the 
sensible  heat  in  the  gases,  plus  the  losses  by  radiation  and 
conduction,  does  not  exceed  10_per  cent,  of  the  calorific  power 
of  the  fuel.  Out  of  the  30  per  cent,  of  the  calorific  power  of  the 
carbon  inevitably  generated,  only  some  10  per  cent,  is,  therefore, 
needed  to  supply  the  losses  in  a  large  producer,  leaving  20  per 
cent,  applicable  to  decomposing  steam.  We  therefore  have: 

Per  1  Kilo,  of  Carbon  Gasified. 

Heat  generated 2,430  Calories. 

Necessary  loss  in  a  large  producer 810 

Useful  for  decomposing  steam 1,620 

Required  to  decompose  1  kilo,  steam  at  212°  F. 

(69,000— 11,466) -5- 18=  3,196 

Maximum  steam  decomposable  1,620-7-3,196  =  0.507  kilo. 

Calculation,  therefore,  shows  that  about  one-half  of  a  unit 
weight  of  steam  is  the  maximum  which  can  be  used  per  unit 


ARTIFICIAL  FURNACE  GAS.  161 

weight  of  carbon  burnt  to  carbon  monoxide,  consistent  with 
keeping  the  producer  at  proper  working  temperature.  This 
would  be  equal  to 

0.088  kilos,  of  steam  per  kilo,  of  air  used. 
0.088  pounds  of  steam  per  pound  of  air  used. 
0.114  kilos,  of  steam  per  cubic  meter  of  air  used. 
0.007  pounds  of  steam  per  cubic  foot  of  air  used. 

The  above  discussion  is  on  the  assumption  that  the  bed  of 
fuel  in  the  producer  is  thick  enough,  and  its  temperature  al- 
ways high  enough,  to  burn  all  the  carbon  to  carbon  monoxide. 
Such  is  only  an  ideal  condition,  for  the  irregular  charging, 
descent  and  working  of  the  fuel  always  allow  of  some  carbon 
dioxide  being  produced,  and  the  best  regular  gas  usually  con- 
tains 1  to  5  per  cent,  of  dioxide,  representing  some  3  to  20  per 
cent,  of  the  total  carbon  oxidized  in  the  producer. 

Assuming  that  on  an  average  10  per  cent,  of  the  carbon 
oxidized  inevitably  forms  carbon  dioxide,  we  can  calculate 
under  these  more  usual  conditions  how  much  steam  can  be. 
used,  because  for  each  kilogram  of  carbon  oxidized,  0.1  kilo, 
then  gives  us  8,100  Calories  per  kilogram  instead  of  2,430  in 
the  producer,  a  surplus  of 

0.1  X  (8,100— 2,430)  =  567  Calories. 

over  the  conditions  when  only  monoxide  is  formed.  Instead 
of  the  1,620  Calories  available  for  decomposing  steam  we  will 
now  have  2,187  Calories,  which  will  decompose 

o  -I  o'-r 

=  0.684  kilo,  of  steam. 

which  reckoned  on  the  air  used  would  be 

0.108  kilos,  of  steam  per  kilo,  of  air  used. 
0.108  pounds  of  steam  per  pound  of  air  used. 
0.140  kilos,  of  steam  per  cubic  meter  of  air  used. 
0.009  pounds  of  steam  per  cubic  foot  of  air  used. 

The  above  proportions  are  those  which  cannot  practically 
be  exceeded,  if  producing  gas  as  low  as  is  usually  possible  in 
carbon  dioxide. 

[Working  the  producer  comparatively  cold,  with  excess  of 


162  METALLURGICAL  CALCULATIONS. 

steam,  much  larger  proportions  of  carbon  dioxide  are  formed 
and  correspondingly  larger  proportions  of  steam  are  decom- 
posed, but  this  manner  of  working  is  abnormal,  is  not  unusual, 
and  will  be  discussed  under  the  next  heading,  treating  of  Mond 
gas.] 

Making  gas  containing  any  given  proportion  of  CO2  to  CO, 
by  volume,  the  ratio  thus  obtained  is  identical  with  the  pro- 
portionate weight  of  carbon  oxidized  to  CO2  and  CO  in  the 
producer,  and  by  the  application  of  the  principles  just  described 
it  can  be  calculated  how  much  steam  can  be  used  per  unit  of 
carbon  oxidized,  making  proper  allowance  for  losses  by  radia- 
tion, etc. 

Illustration'.  A  Siemen's  producer  with  chimney  draft  pro- 
duced gas  containing,  by  volume,  4.  3  per  cent.  CO2  and  25.6 
per  cent.  CO.  How  much  steam  could  be  used  per  pound  of 
air  used,  assuming  50  per  cent,  of  the  heat  generated  by  the 
oxidation  of  the  carbon  to  be  needed  to  run  the  producer? 

Solution : 

4  3 

Per  cent,  of  carbon  burnt  to  CO2  =  =  14.4  per  cent. 

4 .  o  -J~  do .  b 

Heat  generated  by  C  to  CO2  =  0.144X8,100  =1,166  Ib.  Cal. 

Heat  generated  by  C  to  CO     =  0.856X2,430  =  2,080 

Heat  generated  per  kilo,  of  C  =  3,246 

Heat  lost  by  radiation,   etc.  (50  per  cent.)  =  1,623 

Heat  available  for  decomposing  steam  =  1,623 

1  623 
Steam  decomposable  =    '       =  0.508  pounds. 

O  j  A  V/O 

The  above  is  expressed  per  kilogram  of  carbon  oxidized,  but 
the  same  proportion  is  true  per  pound.  The  air  required  per 
pound  of  carbon  oxidized  is  found  from  the  oxygen  required 
to  form  CO  and  CO2: 

Oxygen  for  C  to  CO2  =  0.144x8-3  =    0.384  Ibs. 

Oxygen  for  C  to  CO    =0.856X4-3  =    1.141    " 

Total  =     1.525    ". 

fi  «na  Air  =    6-608    " 

Volume  of  air  =  -  81.8  cu.  feet. 


ARTIFICIAL  FURNACE  GAS.  163 

Steam  per  cubic  foot  of  air  =     '  =     0.006  Ibs. 

ol .  o 

Steam  per  pound  of  air  =     '       •  =    0.077    " 

Air  required  per  pound  of  steam  =  13.0 

It  should  easily  be  seen  that  whatever  heat  is  absorbed  in  the 
producer  in  decomposing  steam,  is  entirely  recovered  when  the 
hydrogen  thus  produced  is  burnt  and  steam  is  reproduced.  If 
then,  in  any  case,  it  is  possible  to  absorb  in  the  producer,  in 
the  decomposition  of  steam,  an  amount  of  heat  equal  to  say, 
20  per  cent,  of  the  total  calorific  power  of  the  fuel,  then  that 
20  per  cent,  is  regained  and  capable  of  being  utilized  when  the 
hydrogen  so  produced  is  burnt  in  the  furnace.  In  other  words, 
20  per  cent,  less  of  the  calorific  power  of  the  fuel  will  be  lost 
in  the  process  of  conversion  into  gas  in  the  producer,  and  20 
per  cent,  more  will  be  obtained  in  the  burning  of  the  gas  when 
it  is  used.  The  great  advantages  of  using  steam  judiciously 
are  thus  clearly  evident. 

Problem  18. 

R.  W.  Hunt  £  Co.  report  the  following  tests  made  of  the 
running  of  a  Morgan  continuous  gas  producer.  Coal  used, 
"  New  Kentucky  "  Illinois  coal,  run  of  mine.  Composition: 

Fixed  carbon 50.87  per  cent. 

Volatile  matter 37.32 

Moisture 5.08 

Ash , 6.73 

100.00 
The  ultimate  composition  was: 

Total  carbon 69.72 

Hydrogen 5.60 

Nitrogen 2.00 

Total  sulphur 0.94 

Oxygen 11.00 

Moisture 5.08 

Inorganic  residue  (less  sulphur) 6.66 

The  ash,  on  combustion,  contains  1.12  per  cent,  of  its  weight 


164  METALLURGICAL  CALCULATIONS. 

•  of  sulphur   (as   FeS);  the  ashes  obtained  from  the  producer 
contain  4.66  per  cent,  of  unburnt  carbon. 

The  gas  produced,  dried,  contained  by  volume: 

Carbon  monoxide  (CO) 24.5  per  cent. 

Marsh  gas  (CH4) 3.6 

Ethylene  (C2H4) 3.2 

Carbon  dioxide  (CO2) 3.7 

Hydrogen  (H2) 17.8 

Oxygen  (O2) 0.4 

Nitrogen  (N2)  (by  difference) 46.8 

[The  moisture  and  sulphur  compounds  in  the  gas  not  having 
been  determined,  we  can  calculate  the  former,  and,  for  the 
purposes  of  calculation,  are  justified  in  assuming  the  sulphur 
present  in  the  gas  as  H2S,  and  in  subtracting  it  from  the  hydro- 
gen. We  will  also  assume  that  the  moisture  in  the  coal  goes 
unchanged  into  the  gases  as  moisture,  and  that  all  the  steam 
used  is  decomposed.  The  0.94  per  cent,  of  sulphur  in  the 
coal  will  furnish  0.86  per  cent,  to  the  gases,  because  6.73  X 
0.0112  =  0.08  per  cent,  will  go  into  the  ash  as  ferrous  sulphide. 
The  0.86  pound  of  sulphur  would  produce  0.91  pounds  of  H2S, 
equal  in  volume  to  9.56  cubic  feet  per  100  pounds  of  coal  used, 
or  (since  we  will  see  later  that  53.83  cubic  feet  of  gas  are  pro- 
duced per  pound  of  coal)  there  will  be  0.2  per  cent,  of  H2S  in 
the  gases,  leaving  17.6  per  cent,  of  hydrogen.] 
On  the  basis  of  above  data  and  assumptions: 
Required:  (1)  The  volume  of  gas  produced  per  pound  of 
fuel  used  in  the  producer. 

(2)  The  weight  of  steam  used  per  100  cubic  feet  of  air  blown 
in,  assuming  the  air  dry. 

(3)  The  proportion  of  the  total  heat  generated  in  the  pro- 
ducer which  is  utilized  in  decomposing  steam. 

(4)  The    percentage    of    increased    economy    thus    obtained 
reckoned  on  the  calorific  power  of  the  fuel. 

(5)  The  efficiency  lost  by  unburnt  carbon  in  the  ashes. 

(6)  The  efficiency  of  the  gas,  burnt  cold,  compared  with  the 
coal  from  which  it  is  made. 

Solution:  (1)  Per  pound  of  coal  burnt,  there  remains  in 
the  ashes  0.0673  X  (0.0466 -i- 0.9534)  =0.0033  pounds  of  un- 
burnt carbon,  leaving  0.06972— 0.0033  =  0.6939  pounds  gasified. 


ARTIFICIAL  FURNACE  GAS.  165 

One  cubic  foot  of  gas,  at  32°  F.,  contains  the  following  weight 
of  carbon: 

C  in  CO     =  0.245X0.54  ounces 
CinCH4    =  0.036X0.54       " 
CinC2H4  =  0.032X1.08       " 
CinCO2    =  0.037X0.54       * 
Total  =  0.382X0.54       " 

=  0.20628  ounces  av 

=  0.01289  pounds  av. 

Gas  produced,  measured  dry,  per  pound  of  coal,  at  32°  F.: 
0.6939 


0.01289 


=  53.83  cubic  feet 


(2)  If  the  moisture  of  the  coal  is  assumed  to  pass  unchanged 
into  the  gas,  as  moisture,  then  all  the  hydrogen  in  the  dry 
gas,  in  any  form  or  combination,  must  have  come  either  from 
hydrogen  in  the  coal  or  in  the  air  blast.  The  hydrogen  in  the 
coal  is  given  as  5.60  per  cent.  The  hydrogen  in  the  gas  is 
calculated  as  follows,  per  cubic  foot: 

H2  in  H2      =  0.176X0.09  ounces. 
H2in  H2S    =  0.002X0.09       " 
H2in  CH4    =  0.036X0.18       " 
H2inC2H4  =  0.032X0.18       " 
Total  =  0.314X0.09       " 

=  0.02826  ounces  av. 

=  0.00177  pounds  av. 

Hydrogen  in  gas  from  1  pound  of  coal: 

0.00177X53.83  =  0.0953  pounds. 
Hydrogen  from  decomposition  of  steam: 

0.0953—0.0560  =  0.0393  pounds. 

Weight  of  steam  decomposed  per  pound  of  coal  used: 
0.0393X9  =  0.3537  pounds. 

To  express  this  weight  relatively  to  the  air  blown  in,  we 
must  calculate  the  air  used  per  pound  of  coal,  as  follows: 


166  METALLURGICAL  CALCULATIONS. 

Nitrogen  in  gas  per  cubic  foot 

0.468X(14X.09)  =  0.5897    oz.  av. 

=  0.036856  Ibs.  av, 

Per  pound  of  coal  =  0.036856X53.83  =  1.9840 

Subtract  N2  in  1  pound  of  coal  =  0.0200 

Leaves  N2  from  air  =  1.9640  " 

1 S 

Weight  of  air  =  1.9640Xy^  =  2.5532 

Volume  of  air  =  2.5532x16^-1.293  =  31.61  cu.  ft. 

Steam  used  per  100  cubic  feet  of  air  blown  in: 

0.3537 


31.61 


X100  =  1.119  pounds.     •  (2) 


(3)  The  heat  utilized  in  decomposing  steam  has  been  found 
to  be  3,196-pound  Calories  per  pound  of  steam  at  212°  F.  We 
therefore,  have  the  heat  so  used  per  pound  of  fuel  used: 

0.3537X3,196  =  1,130-pound  Calories. 

This  quantity  must  now  be  compared  with  the  total  heat  gen- 
erated in  the  producer,  and  the  latter  quantity  can  be  deter- 
minated in  two  ways:  (1)  We  may  subtract  from  the  total 
calorific  power  of  the  coal  the  calorific  power  of  the  gas  pro- 
duced and  of  the  unburnt  carbon  in  the  ashes;  the  difference 
must  be  the  net  heat  generated  in  the  producer,  i.e.,  the  total 
heat  generated  minus  that  absorbed  in  decomposing  steam. 
The  total  heat  generated  is  the  net  heat  thus  calculated  plus 
the  heat  absorbed  in  decomposing  steam.  (2)  We  may  calcu- 
late the  heat  of  formation  of  the  CO  and  CO2  in  the  gas,  and 
assume  that  as  the  total  heat  generated  in  the  producer.  This 
method  is  not  so  accurate  as  (1). 

The  total  calorific  power  of  the  coal  is  given  as  practically 
7,747-pound  Calories  per  pound,  water  formed  being  con- 
densed, which  would  be  decreased  by  the  latent  heat  of  vapor- 
ization, if  the  latter  is  assumed  un condensed.  The  deduction 
is  606.5  X  [(0.056X9) +0.0508]  =  337-pound  Calories,  leaving 
7,410-pound  Calories  as  the  practical  metallurgical  calorific 
power  of  the  fuel. 


ARTIFICIAL  FURNACE  GAS.  167 

Calorific  power  of  the  gas  (dried)  per  cubic  foot: 

CO  0.245 X  3,062  =  750.2  ounce  Cal. 
CH4  0.036  X  8,598  =  309.5 
C2H4  0.032X14,480  =  463.4 
H2  0.176X  2,613  =  459.9 
H2S  0.002X  5,513  =  11.0 
Total  =  1994.0 

=     124.6  pound  Cal. 

Calorific  power  of  gas  per  pound  of  coal: 

124.6X53.83-  6,707  pound  Cal. 

Calorific  power  of  carbon  in  ashes: 

0.0033X8,100= 27 

Sum  =  6,734 

Calorific  power  of  1  pound  coal  =  7,410 

Net  heat  lost  in  conversion  =     676 

Used  in  decomposing  steam  =  1,130 

Gross  heat  generated  in  producer  =  1,806 

Proportion  of  this  utilized  in  decomposing  steam: 

=  0.626  =  62.6  per  cent.  (3) 

(4)  We  can  state  this  result  in  another  way,  by  saying  that 
1,806-7-7,410  =--  24.40  per  cent,  of  the  calorific  power  of  the 
fuel  is  generated  in  the  producer,  of  which   1,130 -h  7, 410  = 
15.25  per  cent,  is  utilized  to  decompose  steam,  and  9.15  per 
cent,  is  lost  by  radiation,  conduction  and  sensible  heat  in  the 
gases.     The   calorific   power   of   the   gases   represents    6,707 -r- 
7,410  =  90.50  per  cent,  of  the  calorific  power  of  the  coal  of 
which  15.25  per  cent.,  however,  is  clear  gain  from  the  employ- 
ment of  steam.     Reckoning  on  the  total  calorific  power  of  the 
coal,  the  increased  economy  from  the  use  of  steam  is  15.2 
per  cent.  (4) 

(5)  The  loss  of  calorific  power  by  the  unburnt  carbon  in 
the  ashes  is  27-pound  Calories,  or,  on  the  whole  heat  available, 

=  7^0  =0.36  per  cent. 


168  METALLURGICAL  CALCULATIONS. 

This  loss  is  exceptionally  low,  and  may  be  very  profitably 
compared  with  the  analogous  loss  of  18.6  per  cent,  occurring 
in  the  case  discussed  in  Problem  16. 

(6)  This  has  already  been  calculated  as 


90.50  per  cent. 

§  ,4-LU 

on  the  assumption  that  the  gases  are  burnt  cold.  If  they  are 
burnt  hot,  say  issuing  from  the  producer  at  1200°  F.  (649°  C.), 
and  are  burnt  when  at  1,000°  F.  (548°  C.)  their  sensible  heat 
at  1,000°  F.  will  be  added  to  their  efficient  heating  power, 
and  can  be  calculated  with  exactness,  using  the  principle  ex- 
plained and  used  in  requirement  (6)  of  Problem  14.  Under 
such  conditions  the  total  efficiency  of  the  producer,  reckoned 
on  the  calorific  power  of  the  coal  used,  approximates  95  per 
cent. 

To  be  fair  to  everybody  concerned,  however,  we  must  de- 
duct from  this  the  coal  required  to  be  burnt  to  raise  the  steam 
used.  There  is  a  little  over  one-third  pound  of  steam  used 
per  pound  of  coal  used  in  the  producer.  This  requires  in 

ordinary  boiler  practice  ~o"X-^-  =  7^7  pound  of  coal,  or  about  4 

per  cent,  of  the  weight  of  fuel  burnt  in  the  producer.  The 
results  of  the  previous  calculation  must,  therefore,  be  dimin- 
ished in  this  proportion,  if  the  steam  has  to  be  raised  by  burning 
coal  under  boilers.  Under  these  conditions  (6)  becomes: 

90.50-^1.04  =  87  per  cent,  efficiency.  (6) 

If  the  steam  can  be  obtained  from  waste  gases  of  a  blast 
furnace,  or  from  the  hot  producer  gases  themselves  (in  case 
they  are  going  to  be  burnt  cold),  then  no  such  deductions  need 
be  made.  It  is  very  evident,  however,  that  if  9.15  per  cent, 
greater  efficiency  is  gained  by  burning  4  per  cent,  more  coal 
to  raise  the  steam,  that  the  net  gain  would  be  only  5.15  per 
cent.  Even  under  these  conditions  it  pays  to  use  steam,  be- 
cause of  the  greater  calorific  intensity  of  the  richer  gas,  the 
usefulness  of  the  steam  for  supplying  air  blast,  and  the  rotting 
of  the  clinkers  therewith  obtained. 


ARTIFICIAL  FURNACE  GAS.  169 

3.     MOND  GAS. 

In  the  Mond  producer,  an  excess  of  steam  is  introduced  with 
the  heated  air  used  for  combustion;  the  producer  is  thus  run 
much  colder  than  the  ordinary  producer,  and  far  more  carbon 
dioxide  is  present  in  the  gas.  In  fact,  the  gas,  compared  with 
ordinary  producer  gas,  is  very  high  in  carbon  dioxide  (10  to 
20  per  cent.),  very  high  in  hydrogen  (20  to  30  per  cent.),  very 
low  in  carbon  monoxide  (10  to  15  per  cent.),  low  in  nitrogen 
(40  to  50  per  cent.),  and  carries  an  extraordinary  amount  of 
undecomposed  moisture.  The  fuel  used  is  low-grade  bitumin- 
ous slack,  and  the  object  of  using  so  much  steam  is  to  keep 
the  temperature  in  the  producer  so  low  that  a  maximum  amount 
of  the  nitrogen  in  the  coal  is  evolved  as  ammonia.  The  gas 
is  cooled  to  ordinary  temperature  by  contact  with  water  spray, 
so  that  all  but  a  small  amount  of  moisture  is  condensed,  the 
ammonia  is  removed  by  dilute  sulphuric  acid,  and  the  cold 
nearly  dry  gas  is  then  used  for  gas  engines  or  in  furnaces. 

The  calorific  power  of  the  gas  is  not  low,  because  the  high 
proportion  of  hydrogen  compensates  for  the  low  carbon  mon- 
oxide, while  the  great  heat  evolved  by  the  large  formation  of 
carbon  dioxide  has  been  mostly  absorbed  in  decomposing 
steam,  and  is,  therefore,  potentially  present  in  the  gas  in  the 
form  of  hydrogen. 

Problem  19. 

Bituminous  slack  coal  used  in  Mond  producers  for  gener- 
ating gas  for  a  gas-engine  power  plant  contained: 

Moisture 8.60  per  cent. 

Carbon..  . ..62.69 

Hydrogen 4.57 

Oxygen . .  .10.89 

Nitrogen 1.40 

Ash 10.42 

Calorific  power,  determined  in  a  bomb  calorimeter,  water 
condensed,  6,786  Calories  per  unit  of  dried  fuel.  Ashes  pro- 
duced 268  pounds  per  ton  (2,240  pounds)  of  moist  slack  used; 
contains  12  per  cent,  of  carbon. 

Air  used  for  running  is  heated  to  300°  C.  by  the  waste  heat 
of  producer  gases,  and  carries  in  2J  tons  of  water  as  steam  (at 


170  METALLURGICAL  CALCULATIONS. 

same  temperature)  for  every  ton  of  fuel  burnt.  The  steam 
is  generated  by  a  tubular  boiler  run  by  the  escape  gases  from 
the  gas  engines,  but  is  heated  from  100°  C.  to  300°  C.  by  the 
waste  heat  of  the  producer  gases.  The  latter  escape  from  the 
producer  at  350°  C. 

Composition  of  waste   gases   passing  out   of   condensers  at 
15°  C.: 

Carbon  monoxide  (CO) 11.0  per  cent 

Hydrogen  (H2) 27.5 

Marsh  gas  (CH4) 2.0 

Carbonic  oxide  (CO2) 16.5 

Nitrogen  (N2) 41.3 

Water  vapor  (H20) 1.7        " 

100.0 
.     Assume  all  the  nitrogen  of  the  fuel  to  form  ammonia  gas  NH3. 

Required: 

(1)  The  calorific  power  of  the  Mond  gas. 

(2)  The  volume  of  gas  produced  per  ton  of  fuel  used. 

(3)  The  efficiency  of  the  producer.      , 

(4)  The  weight  of  steam  which  is  decomposed  in  the  pro- 

ducer. 

(5)  The  proportion  of  the  calorific  power  of  the  fuel  saved 

to  the  gas  by  the  decomposition  of  steam. 

(6)  The  proportion  of  the  heat  generated  in  the  producer 

which  is  saved  to  the  gas  by  the  decomposition  of 
steam. 

(7)  The  proportion  of  the  calorific  power  of  the  coal  lost 

from  the  producer  by  radiation  and  conduction. 

Solution:    (1)  One  cubic  foot  of  the  gas  at  0°  C.  would  gen- 
erate the  following  amounts  of  heat  (water  un condensed). 

CO  =  0.110  cubic  feet  X  3062  =   336.8  ounce  Calories. 
H2  =  0.275  cubic  feet  X  2613  =   718.6 
CH4  =  0.020  cubic  feet  X  8598  =    172.0 

Total  =1227.4 

=     76.7  pound 
=    138.1  B.  T.  U. 
Per  cubic  meter   =  1227.4  kilo. 


ARTIFICIAL  FURNACE  GAS.  171 

If  measured  at  15°  C.  (60°  F.)  the  above  values  will  be  re- 
duced by  the  factor  273  -r  (273  +  15),  and  become 

Per  cubic  foot     =     72.7  pound  Calories. 

Per  cubic  meter  =  1163  kilo.         "  (1) 

(2)  Carbon  in  1  cubic  foot  of  gaa  at  0°  C.: 
In  CO    0.110X0.54  ounces 
In  CH4  0.020X0.54       " 
In  CO2  0.165X0.54       " 

0.295X0.54       "      =  0.1593  ounces 

=  0.009956  pounds 

Carbon  going  into  gas  per  pound  of  fuel  burnt: 
Carbon  in  fuel  0.6269  pounds 


Carbon  in  ashes  ~-X0.12  =  0.0144 

Carbon  in  gas  =  0.6125       " 

Volume  of  gas  (at  0°  C.)  per  pound  of  fuel  used: 

0.6125 

=  61.52  cubic  feet 


0.009956 

Per  ton  of  2240  pounds  =  137,805  cubic  feet 

At  15°  C.  (60°  F.) 

=  137,805  X27;!*15  ==  145,375     "    "  (2) 


(3)  The  calorific  power  of  the  dried  fuel  is  given  as  6,786 
Calories  per  pound,  water  condensed.  Since  one  pound  of 
wet  fuel  contains  100—8.60  =  91.40  per  cent,  dried  fuel,  the 
calorific  power  of  1  pound  of  wet  fuel,  moisture  condensed,  is 

6,786X0.9140  =  6,202  pound  Calories. 
But,  1  pound  of  wet  fuel  would  produce,  on  combustion, 

0.0860  +  9  (0.0457)  =  0.4973  pounds  moisture, 
which,  remaining  vaporized  at  15°,  would  retain 

0.4973X596  =  296  Calories, 
leaving  the  net  metallurgical  calorific  power  as 

6,202—296  =  5,906  Calories. 


172  METALLURGICAL  CALCULATIONS. 

The  61.52  cubic  feet  of  gas  produced  per  pound  of  moist  fuel 
will  have  a  calorific  power,  burnt  cold,  of 

61.52X76.7  =  4,719  Calories, 
making  the  efficiency  of  the  producer,  on  fuel  consumed  in  it, 

4  71Q 

g^-  0.799  =  79.9  per  cent.  (3) 

The  above  figure  is  true  only  on  the  assumption  that  the 
steam  used  is  obtained  from  waste  heat,  and  therefore  does 
not  require  the  combustion  of  extra  fuel 

(4)  The  gas  produced  contains,  per  cubic  foot,  the  following 
amount  of  hydrogen,  free  and  as  CH4: 

As  H2    0.275  cubic  feet  X  0.09  =0.02475    ounces. 

As  CH4  0.040  cubic  feet  X  0.09  =  0.00360 

Sum  =  0.02835 

=  0.001 772  pounds 

Per  pound  moist  fuel  =  0.001772X61.52        =  0.1090 
Present  as  ammonia  gas  0.0140  X  (3  H-  14)        =  0.0030 
Total  (not  including  H  as  water)  =  0.1120 

Hydrogen  in  1  pound  of  coal  =  0.0457  " 

Hydrogen  from  decomposition  of  steam          =  0.0663  " 

Water  decomposed  in  the  producer  =  0.5967  "(4) 

Since  the  steam  introduced  weighs  2.5  pounds  for  every 
pound  of  fuel  burnt,  we  see  that  only 

0.2387  =  23.87  per  cent. 

j . 

of  the  steam  introduced  is  decomposed. 

[In  the  writer's  opinion  this  unused  76.85  per  cent,  can  only 
pass  in  and  pass  out  carrying  out  sensible  heat,  and  it  seems 
a  very  wasteful  method  of  keeping  down  the  temperature  in 
the  producer.  From  the  standpoint  of  regarding  the  2  pounds 
of  steam  as  a  mere  absorber  of  sensible  heat,  it  could  probably 
be  replaced  by  some  of  the  gases  of  combustion  from  the  gas 
engine  or  open-hearth  furnace.  The  products  of  combustion 
of  the  above  gas  would  contain  approximately 

Nitrogen 68.7  per  cent. 

Carbon  dioxide 14.7 

Water  vapor 16.6 


ARTIFICIAL  FURNACE  GAS.  173 

And  if  the  J  pound  of  water  vapor  decomposed  in  the  pro- 
ducer were  thus  supplied,  it  would  bring  in,  per  pound  of  coal 
burnt, 

Nitrogen  .........................  41.2  cubic  feet. 

Carbon  dioxide  .................   8.8 

Water  vapor  ...................   9.9 

And  if  the  CO2  thus  introduced  were  reduced  to  CO,  as  it  prob- 
ably would  be,  the  gases  would  receive  from  this  source 

Nitrogen  .......................  41.2  cubic  feet. 

Carbon  monoxide  ...............  27.5 

Hydrogen  ......................   9.9 

thus  producing  gas  quite  up  to  standard  as  regards  combus- 
tibles, while  the  heat  absorbed  in  the  reduction  of  CO2  to  CO 
would  cool  the  fire  down  quite  as  effectually  as  the  extra  steam 
now  used.] 

(5)  The  steam  decomposed,  0.5967  pounds  per  pound  of  fuel 
burnt,  may  be  assumed  to  be  at  100°  C.  on  entering  the  fire, 
and  to  therefore  absorb  3,196-pound  Calories  per  pound  of 
steam  decomposed  '.  The  heat  absorbed  in  the  producer,  and 
thus  transferred  into  potential  calorific  power  is,  therefore 

3196X0.5967  =  1907  pound  Calories, 
which  expressed  in  per  cent,  of  the  calorific  power  of  the  fuel    is 

=  0.323  =  32.3  per  cent  (5) 


(6)  The  heat  generated  in  the  producer  equals  the  calorific 
power  of  the  coal  minus  the  heat  lost  by  carbon  in  the  ashes, 
minus  the  calorific  power  of  the  gases,  plus  the  heat  absorbed 
in  decomposing  steam.  The  loss  by  carbon  in  the  ashes  is 

X0.12X8100  =  117  Calories. 


2240 
The  heat  generated  in  the  producer  is,  therefore, 

5906—117 — 1719  +  1850  =  2920  Calories, 
equal  to  2920  -r-  5906  =  49.4  per  cent,  of  the  calorific  power  of 


174  METALLURGICAL  CALCULATIONS. 

the  coal.     Of  this,  1907  Calories  is  absorbed  in  decomposing 

1907 

steam,  which  is  TTQ™  =  65.3  per  cent,  of  the  total  heat  gener- 


ated. (6) 

(7)  There  are  2922  Calories  generated  in  the  producer,  of 
which  1907  are  absorbed  in  decomposing  steam,  leaving  1015 
Calories  to  supply  radiation  and  conduction  and  as  sensible 
heat  in  the  hot  gases,  to  which  must  be  added  the  sensible 
heat  in  the  hot  air  and  steam  used  at  300°  C. 

The  air  used  per  pound  of  coal  is  found  from  the  nitrogen 
in  the  gases: 

Nitrogen  in  1  cubic  foot  of  producer  gas  =  0.413  cubic  foot 

Nitrogen  in  61.52  cubic  foot  of  producer  gas  =  25.40 

Air  used  =  25.4-0.792  =32.08 

Steam  used  =  (2.5X16)^0.81  =  49.40 

Heat  in  steam  and  air  at  300°  C.  (from  15°  C.): 

32.08X0.3116X285  =  2850  ounce  Calories 
49.40X0.3872X285=5450 


Sum  =  8300 

=    519  pound  Calories 

Total  heat  radiated,  conducted  and  in  hot  gases: 
1015  +  519  =  1534  Calories. 

The  heat  in  the  hot  gases  is  as  follows,  per  cubic  foot  of  gas 
produced : 

CO   0.110X0.313X335] 

H2    0.275X0.313X335  [    =    83.7 

N2    0.413X0.313X335  J 

CO2  0.165X0.450X335      =    24.9 

CH4  0.020  X  0.460  X  335       =      3. 1 

Sum      =  111.7  ounce  Calories 
=      7.0  pound  Calories. 
Per  pound  of  coal  burnt : 

7.0X61.52  =  430  pound  Calories 


ARTIFICIAL  FURNACE  GAS  175 

To  this  must  be  added  the  heat  'in  undecomposed  water  vapor 
in  the  gases,  as  follows: 

Steam  used  per  pound  of  coal  =     2.50    pounds 

Steam  decomposed  per  pound  of  coal  =    0.60 
Steam  remaining  in  gases  =     1.90 

Moisture  in  coal  =    0.086 

Total  in  gases  =    1.986 

Volume  =  (1.986X16)-*-  0.81  =  39.2     cubic  feet 

Heat  contained  in  this  at  350°  C.  : 

39.2X0.395X335  =  5187  ounce  Calories 
=    324  pound  Calories 

Total  heat  in  producer  gases: 

430  +  324  =  754  Calories 
Heat  lost  by  radiation  and  conduction: 

1589—754  =  827  Calories 
Proportion  of  calorific  power  of  coal  thus  lost: 


=  0.140  =  14.0  per  cent.  (7) 


When  Mond  gas  is  heated  in  the  regenerator  of  an  open- 
hearth  furnace  it  changes  considerably  in  composition,  as  is 
shown  by  the  following  analyses  made  by  Mr.  J.  H.  Darby,  on 
gas  dried  before  analysis: 

Before  After 

Regenerator.  Regenerator. 

Carbonic  acid  gas,  CO2 17.8  10.5 

Carbon  monoxide,  CO 10.5  21.6 

Ethylene,  C2H4 0.7  0.4 

Methane,  CH4 2.6  2.0 

Hydrogen,  H2 24.8  17.7 

Nitrogen,  N2 43.6  47.8 

100.0  100.0 

The  above  changes  are  very  interesting,  and  their  discussion 
profitable.  Before  heating,  the  gas  burns  with  a  non-luminous 


176  METALLURGICAL  CALCULATIONS. 

flame;  after  heating,   it  burns  with  a   brilliant  white   flame. 
Let  us  inquire: 

(1)  What  chemical  change  occurs  during  the  heating? 

(2)  What  are  the  relative  volumes  of  the  gas  before  and 
after  heating  (excluding  water)? 

(3)  What  change  in  the  calorific  power  is  produced  by  the 
heating  ? 

(1)  An  inspection  of  the  analyses  shows  undoubtedly  that 
at  a  high  temperature  the  CO2  cannot  hold  all  its  oxygen  in  the 
presence  of  such  a  large  amount  of  hydrogen,  and  that  the 
following  reaction  must  occur: 

i         i          i         i 
C02  +  H2  =  CO  +  H2O 


The  figures  given  do  not  check  exactly,  but  assuming  that 
the  above  reaction  takes  place,  we  can  find  to  what  extent,  by 
expressing  the  composition  of  the  gas  after  heating  for  the 
same  amount  of  nitrogen  as  was  in  the  gas  before  heating, 
since  this  gas  is  unchanged: 

Before  After  Loss 

Heating.  Heating.  or  Gain. 

CO2  ....................  17.8  9.6  —8.2 

CO  .....................  10.5  19.7  +9.2 

H2  ................  .'....24.8  16.1  —8.7 

N2  .....................  43.6  43.6  0.0 

Since,  according  to  the  reaction  written,  the  volume  of  CO2 
reduced  to  CO  will  be  equal  to  the  volume  of  H2  thus  con- 
sumed, and  will  produce  an  equal  volume  of  CO;  the  above 
table  proves,  within  the  probable  limits  of  error,  that  the 
reaction  written  actually  takes  place. 

The  separation  of  luminous  carbon  is  probably  due  to  the 
splitting  up  of  C2H4. 

The  relative  volumes  of  the  heated  and  unheated  gases  will 
be  inversely  as  the  percentage  of  nitrogen  in  each  (since  this 
gas  is  unchanged),  viz.:  as  47.8  to  43.6,  or  as  100  to  91.2.  The 
contraction,  8.8  parts,  would  again  correspond  almost  exactly 
to  the  amount  of  water  formed  in  the  assumed  reaction,  which 
would  be  equal  to  the  hydrogen  ^so  used,  thus  giving  another 
check  on  the  validity  of  the  reaction  assumed  to  take  place. 


ARTIFICIAL  FURNACE  GAS.  177 

The  calorific  power  of  1  cubic  foot  of  original  gas  is: 
CO     0.105X   3,062=    321.5  ounce  Calories 
C2H4  0.007X14,480  =     101.4      " 
CH4    0.026  X   8,598  =    223.5      " 
H2      0.248  X   2,613  =    648.0      " 
Sum  =  1294.4      " 

=      80.9  pound      " 
Per  100  cubic  feet     =     8090       " 

The  calorific  power  of  1  cubic  foot  of  heated  gas  is: 
CO     0.2I6X   3,062=    661.4  ounce  Calories 
C2H4  0.004X14,480  -      57.9      " 
CH4   0.020  X   8,598  =     172.0      " 
H2      0.177X   2,613  =    462.5      " 
Sum  =  1353.8      " 

=      84.6  pound 

For  91.2  cubic  feet  the  calorific  power  would  be  84.6X91.2 
=  7715  pound  Calories. 

The  net  conclusion  is  that  the  heated  gas  (aside  from  its 
sensible  heat)  gives  less  heat  by  combustion  in  the  furnace 
per  unit  of  coal  gasified  in  the  producers  than  the  correspond- 
ing quantity  of  unheated  gas;  but  the  difference  is  only  some  4 
per  cent.  On  the  other  hand,  the  calorific  power  of  the  heated 
gas  per  cubic  foot  is  some  5  per  cent,  greater  than  that  of  the 
unheated  gas,  and,  therefore,  its  calorific  intensity  will  have  been 
increased  by  the  heating — quite  aside  from  the  question  of  its 
temperature  being  higher,  and  therefore  its  sensible  heat  greater. 

4.  WATER  GAS. 

This  gas  is  made  intermittently,  by  first  burning  part  of  the 
fuel  until  the  fire  is  very  hot,  and  then  introducing  steam, 
cool  or  superheated,  into  the  fire.  The  reaction  producing  the 
gas  is: 

i  ii 

C  +  H20  =  CO  +  H2 

And  if  the  reaction  is  complete,  and  only  carbon  is  present  as 
fuel,  the  gas  produced  is  theoretically  composed  of  equal  parts 
by  volume  of  CO  and  H2,  the  volume  of  each  of  these  being 
equal  to  that  of  the  steam  used  (if  measured  at  the  same  tem- 
perature and  pressure). 


1/8  METALLURGICAL  CALCULATIONS. 

Deviations  from  this  ideal  composition  are  caused  in  practice 
by  the  occurrence  of  undecomposed  steam  in  the  gas,  also  of 
CO2  and  N2,  which  come  from  the  residual  gas  formed  during 
the  heating  up  of  the  fire,  some  of  which  will  get  into  the  first 
portion  of  the  water-gas  produced,  and  also  hydrocarbons,  tar 
and  ammonia  from  the  distillation  of  the  fuel.  A  typical 
analysis  of  water-gas,  given  by  Mr.  W.  E.  Case,  is: 

Hydrogen  (H2) 48.0 

Carbon  monoxide  (CO) 38.0 

Methane  (CH4) .   2.0 

Carbon  dioxide  (CO2) 6.0 

Nitrogen  (N2) 5.5 

Oxygen  (O2) 0.5 

The  production  of  water-gas  is  more  expensive  than  that  of 
other  artificial  producer  gases,  because  of  the  large  amount 
of  steam  necessarily  required,  the  heat  lost  during  the  heating 
up,  and  the  essentially  intermittent  character  of  the  operation. 
Its  essential  advantages  are  its  very  high  proportion  of  com- 
bustibles, averaging  90  per  cent,  and  the  consequent  high 
calorific  intensity  which  it  is  capable  of  producing.  (Uncar- 
buretted,  it  is  well  known  that  water-gas  is  a  valuable  domestic 
fuel,  and  illuminant  when  used  in  mantle  burners;  carburetted, 
it  forms  our  principal  illuminating  gas,  and  as  such  is  manu- 
factured on  an  immense  scale.  We  will  treat  here  only  of  its 
metallurgical  uses.) 

Discussing  the  manufacture  of  the  gas,  the  first  step  is  the 
heating  up  of  the  fuel.  This  is  accomplished  by  blowing  air 
through  it.  At  this  point  we  can  distinguish  two  systems. 
The  older  one  is  that  of  blowing  in  air  under  moderate  pres- 
sure, which  passes  through  the  fire  at  moderate  velocity,  and 
produces  a  fair  grade  of  ordinary  producer  gas.  This  gas  is 
either  wasted,  or  burned  in  furnaces  requiring  such  quality  of 
gas,  or  burned  in  regenerators  where  heat  is  stored  up  to  be 
utilized  in  the  next  stage  in  superheating  steam.  The  newer 
system  is  to  blow  in  air  at  high  pressure,  such  that  a  large 
percentage  of  carbon  dioxide  remains  in  the  gases,  thus  nearly 
completely  burning  what  carbon  is  oxidized,  and  storing  up  a 
corresponding  quantity  of  heat  in  the  remaining  carbon.  The 
incombustible  gases  produced  are  passed  through  a  recuperator 


ARTIFICIAL  FURNACE  GAS.  179 

or  regenerator,  where  their  sensible  heat  is  partly  communi- 
cated to  the  steam  used,  so  as  to  superheat  it  when  forming 
water-gas.  This  system  consumes  the  minimum  of  carbon 
in  "  heating  up  "  the  fuel,  saves  time  in  this  unproductive  period, 
and  allows  the  producer  to  be  run  longer  "  on  steam."  In 
practice  the  fuel  is  probably  raised  to  an  average  temperature 
of  1500°  C.  during  the  heating  up. 

We  can  calculate  the  efficiency  of  this  heating  up,  that  is, 
the  proportion  of  the  calorific  power  of  the  carbon  burnt  which 
is  stored  up  in  the  remaining  fuel,  if  we  know  how  much  fuel 
is  in  the  producer,  how  much  air  is  blown  through,  the  com- 
position of  the  gases  produced,  and  the  average  rise  in  tem- 
perature of  the  fuel.  Since  this  operation  is,  however,  only 
supplementary  to  the  real  formation  of  water-gas  by  the  de- 
composition of  steam,  we  will  first  make  calculations  upon 
the  cooling  down  period,  during  which  water-gas  is  made. 

During  the  use  of  steam  the  reaction  absorbs  heat,  and  the 
producer  rapidly  cools.  Steam  is  passed  through  until  the 
temperature  of  the  fuel  is  800°  C.,  and  must  then  be  stopped, 
because  between  800°  and  600°  the  reaction  is  mostly 

C  +  2H20  =  CO2  +  2H2 

resulting  in  a  rapid  increase  of  CO2  in  the  gas.  The  heat  to 
decompose  steam  is  furnished  partly  by  the  oxidation  of  carbon 
to  CO,  and  partly  by  the  sensible  heat  in  the  carbon  itself. 
Since  the  carbon,  from  temperatures  of  about  1000°  C.  up  has 
a  specific  heat  of  0.5,  we  can  easily  calculate  how  much  steam 
can  be  decomposed  before  the  temperature  falls  to  800°. 

Problem  20. 

A  Dellwick- Fleischer  water-gas  producer  contains  3  tons  of 
coke  (90  per  cent,  carbon),  heated  up  to  1500°  C.  Steam, 
heated  to  300°  C.,  is  passed  through  for  8  minutes,  until  the 
temperature  of  the  gas  escaping  is  700°  C.,  or  800°  at  the  fuel 
bed.  The  composition  of  the  gas  is  (Prof.  V.  B.  Lewes): 

Hydrogen 50.0 

Carbon  monoxide 40 . 0 

Carbon  dioxide 5.0 

Oxygen 1.0 

Nitrogen 4.0 


180  METALLURGICAL  CALCULATIONS. 

Assume  9000  Calories  per  minute  lost  by  radiation  and  con- 
duction. 

Required:  (1)  The  amount  of  steam  used  in  the  8  minutes 
(2)  The  volume  of  gas  produced  in  the  8  minutes. 

(1)  The  amount  of  steam  which  could  have  been  used  is 
limited  by  the  available  heat  to  decompose  it.  The  latter  is 
furnished  by 

(a)  Oxidation  of  carbon  to  monoxide. 
(6).  Oxidation  of  carbon  to  dioxide. 

(c)  Sensible  heat  of  the  carbon  and  ash  of  fuel. 

(d)  Sensible  heat  of  the  steam  used. 

While  the  items  of  heat  absorption  and  loss  are: 

(e)  Decomposition  of  the  steam. 
(/)    Sensible  heat  in  the  gases. 
(g)  Radiation  and  conduction. 

The  simplest  way  to  arrive  at  a  solution  is  to  make  a  few 
necessary  assumptions,  and  to  then  let  X  represent  the  weight 
of  steam  used.  The  assumptions  are  that  the  average  tem- 
perature of  the  fuel  bed  falls  to  1000°  C.,  that  the  average 
temperature  of  the  escaping  gases  is  (1500  +  700)  -=-2  =••  1100°  C., 
that  the  average  specific  heat  of  carbon  in  the  range  1000 — 1500 
is  0.5,  and  of  the  ash  of  the  fuel  0.25.  Then,  casting  up  a 
heat  balance  sheet  in  terms  of  X,  we  can  finally  arrive  at  an 
expression  for  the  heat  which  was  available  during  the  8  minutes 
for  decomposing  steam,  and  thus  at  the  weight  of  steam  decom- 
posed, and  (making  this  equal  to  X)  at  a  solution. 

(a)  The  analysis  of  the  gas  shows  that  eight  times  as  much 
carbon  was  burnt  to  monoxide  as  to  dioxide,  making  the  equa- 
tion of  combustion: 

9C+10H2O  =  8CO  +  C02+10H2 

By  weight,  8X12  =  96  parts  of  carbon  was  burnt  to  CO  per 
10  X 18=  180  parts  of  steam  used,  or  0.533  parts  per  one  part  of 
steam.  The  heat  generated  in  forming  monoxide  is,  therefore : 

0.533XXX2430  =  1296  X  Calories. 

(6)  Only  one-eighth  as  much  carbon  burns  to  dioxide,  giving, 
therefore,  as  the  heat  evolved: 

(0.533H-8)XXX8100  =  540  X  Calories. 


ARTIFICIAL  FURNACE  GAS.  181 

(c)  3000  kilos,  of  fuel,  representing  2700  kilos,  of  carbon  and 
300  kilos,  of  ash,  cool  from  1500°  to  1000°: 

2700X0.50X500  =  675,000  Calories 
300X0.25X500  =    37,500 
Sum  =  712,500 

(d)  X-J-0.81  will  be  the  volume  of  the  steam  used  at  normal 
conditions,  which  brings  in  considered  only  as  vapor,  at   300°: 

(X  •*•  0.81)  X  0.385X300  =  142.6  X  Calories. 

The  sum  total  of  (a)  +  (b)  +  (c)  +  (d)  gives  the  total  available 
heat,  viz.: 

1978.6  X-f  712,500  Calories. 

(e)  The    steam    used    requires    for   its    decomposition,    con- 
sidered theoretically  as  cold  steam,  producing  cold  products: 

(X  ^  9)  X  29,042  =  3227  X  Calories 

(/)  The  gas  being  50  per  cent,  hydrogen,  and  the  latter  being 
equal  to  the  volume  of  steam  used,  the  volume  of  gas  must  be 
2X  (X-:- 0.81),  which  multiplied  by  the  mean  specific  heat  of 
gas  of  this  composition  between  0°  and  1100°  per  cubic  meter, 
and  by  the  temperature,  will  give  the  heat  thus  carried  out  of 
the  producer: 

2  (X-r-0.81)X0.347XllOO  =  942.5  X  Calories 
(g)  9000X8  =  72,000  Calories. 

The  sum  total  of  (e)  +  (f)  -f  (g)  gives  the  total  heat  distribu- 
tion, viz.: 

4.169.5  X  + 72,000  Calories. 

Since  the  heat  available  equals  the  heat  distributed: 

1.978.6  X-f  712,500  =  4,169.5  X 4- 72,000 

or  X  =  292  kilograms  (1) 

(2)  The  volume  of  this  steam,  at  assumed  standard  condi- 
tions, would  be 

292  H- 0.81  =  360  cubic  meters, 
and  of  gas,  since  it  is  50  per  cent,   hydrogen, 
360X2  =  720  cubic  meters. 


182  METALLURGICAL  CALCULATIONS. 

The  above  figures  represent  the  maximum  attainable  pro- 
duction, on  the  assumption  that  sufficient  steam-generating 
power  is  available  to  furnish  the  steam,  and  that  the  fuel  in 
the  producer  is  of  small  size  and  the  bed  so  uniform  that  the 
production  of  gas  is  regular  all  over  it.  In  practice,  figures  con- 
siderably below  this  are  attained,  but  it  is  always  well  to  know 
the  possible  maximum  which  is  attainable. 

Problem  21. 

In  a  Dellwick-Fleischer  water-gas  producer  the  heating  up  is 
accompliched  in  2  minutes  by  blast  from  a  Root  blower,  fur- 
nishing air  through  a  9.5  in  h  pipe  at  a  total  water-gauge 
pressure  of  19  inches  of. water,  temperature  of  air  15°  C.  The 
gases  escaping  from  the  producer  analyze: 

Carbon  dioxide 17.9  per  cent. 

Carbon  monoxide 1.8 

Nitrogen 78.6         " 

Oxygen 1.7 

Temperature  of  waste  gases  900°  C. ;  heat  lost  by  radiation 
and  conduction  9000  Calories  per  minute;  assume  producer  to 
contain  3000  kilos,  of  fuel,  consisting  of  90  per  cent,  carbon  and 
10  per  cent.  ash. 

Required:   (1)  The  average  rise  in  temperature  of  the  fuel  bed. 

(2)  The   proportion   of   the   heat   generated   which  is  thus 
stored  up  as  useful  heat  for  producing  water-gas. 

(3)  Assuming  that  the  production  of  water-gas  lasts  8  min- 
utes, during  which  2500  Calories  are  absorbed  from  the  fuel 
bed  per  kilogram  of  steam  used,  what  should  be  the  steam 
supply  in  kilograms  per  minute? 

(4)  The  ratio  between  the  volume  of  air  supplied  during  the 
blowing  up  period  and  the  weight  of  steam  used  in  the  gas- 
making  period. 

Solution:  (1)  We  imist  first  find  the  amount  of  air  fur- 
nished by  the  blower.  To  do  this,  we  calculate  the  pressure 
head  in  terms  of  air  at  15°  C.,  instead  of  water,  and  then  apply 
the  well-known  formula  V  —  V  2g.  h.  "^ater  is  772  times  as 
heavy  as  air  at  0°  C.,  and,  therefore,  1^  inches  of  water  pres- 
sure would  represent  19X772+12  =  1222  fee'  of  air  pressure 
(that  is,  a  column  of  fluid  as  light  as  air,  1222  feet  high).  But 


ARTIFICIAL  FURNACE  GAS.  183 

air  at  15°  is  lighter  still  than  air  at  0°,  in  the  ratio  273  to  288. 
so  that  the  air  pressure  measured  in  terms  of  air  at  15°  will  be 


=  1290  feet 


The  velocity  of  the  air  supplied  will  therefore  be,  in  feet  per 
second  :  _ 

V  =V  64X1290 

=  287  feet  per  second, 
and  the  volume  delivered,  per  minute,  at  15°  C.: 

287  X  60  X  (0.7854  X  9.5  X  (  .5  -5-  144)  cubic  feet 

=  17,220X0.492  =  8472  cubic  feet, 
which  in  terms  of  air  at  0°  C.,  would  be 

97*3 

847  X  Sss  =  8050  cubic  feet 

Zoo 

=     228  cubic  meters 

The  volume  of  waste  gases  produced  in  the  2  minutes  can 
be  found  from  the  relative  percentages  of  nitrogen  in  the  air 
(79.  2^  and  in  the  gases  (78.6),  as  follows: 

79  9 

228X2X;r~-^  =  459.5  cubic  meters 
/o.  o 

Containing,  therefore,  from  its  analysis: 

Carbon  dioxide  ..............   82.25  cubic  meters 

Carbon  monoxide  ............     8.27      "  " 

Oxygen  .....................     7.81      " 

Nitrogen  ....................  361.15      " 

459.48      « 
The  carbon  burnt  to  CO2  and  CO  will  be: 

C  to  CO2  82.25X0.54  -    44.42  kilos. 
CtoCO     8.27X0.54=     4.47     " 
Sum  =   48.89     " 

And  the  heat  thus  generated: 

C  to  CO2  -  44.42X8100  =  359,800  Calories 
CtoCO    =    4.47X2430  =_JA860 
Sum  =  370,660 


184  METALLURGICAL  CALCULATIONS. 

To  find  the  amount  of  this  heat  left  in  the  producer  at  the 
end  of  the  2  minutes  blowing  up,  we  must  subtract  the  2X 
9000  =  18,000  Calories  lost  by  radiation  and  conduction,  and 
then,  in  addition,  the  heat  carried  out  by  the  hot  gases,  at  an 
average  temperature  of  900°  C.,  which  latter  will  be: 

CO,  O2,  N2  377.25X0.327X885    =  109,150  Calories 
CO2    82.25X0.571X885    =    41,565 

Sum    =  150,715     *  * 

Heat  left  in  the  fuel  bed: 

370,660—168,715  =  201,945  Calories 
heat  capacity  of  the  fuel  bed  per  1°  C.: 

3000X0.9  kilos,  carbon  X0.5    =  1370  Calories 

3QOOX0.1  kilos,  ash         X0.25  = 75 

Sum  =  1445 

Average  rise  in  temperature  of  the  fuel  bed. 

uo-c.  a) 

(2)  The  useful  heat  thus  stored  up  in  the  fuel  bed  amounts 
to  the  following  proportion  of  the  total  heat  generated  during 
the  blowing  up: 

=  0.545  =  54.5  per  cent.  (2) 

(3)  Steam  which  can  be  decomposed  in  the  gas-producing 
period: 

=  80.8  kilograms 

=  10.1  kilograms  per  minute         (3) 

(4)  Air  supplied  in  2  minutes  =  456  cubic  meters.     Steam 
used  in  8  minutes  =  80.8  kilograms. 

80  8 

Ratio  =  -TVTT  =  0.177  kilos,  steam  per  1  m8  air  (4) 

4ob 

=  0.177  ounce  steam  per  1  ft3  air 
—  1.1  pounds  steam  per   100  ft3  air 


CHAPTER  VII. 
CHIMNEY  DRAFT  AND  FORCED  DRAFT. 

In  all  problems  concerning  combustion,  we  must  furnish  the 
air  needed  for  combustion  either  by  suction  or  by  pressure. 
The  original  and  almost  universal  method  is  by  chimney  draft; 
the  more  positive  and  reliable  method  is  forced  draft.  Often 
the  two  are  combined  with  very  satisfactory  results. 

The  waste  heat  from  any  metallurgical  process  or  furnace 
is  generally  considerable.  Most  furnaces  must  be  kept  above 
a  red  heat,  and  the  gases  pass  directly  out  of  the  furnace  into 
the  chimney.  In  such  cases  the  chimney  is  indicated  as  the 
proper  source  of  draft,  because  it  utilizes,  although  very  in- 
efficiently, the  ascensive  force  of  the  hot  gases,  and  thus  works 
by  otherwise  wasted  energy.  In  other  cases  it  is  practicable 
to  pass  the  gases  through  boilers  before  they  go  to  the  chim- 
ney, and  thus  to  raise  large  amounts  of  steam.  The  gases  are 
then  cooled  down  so  far  that  they  enter  the  chimney  too  cold 
to  furnish  all  the  draft  needed;  in  such  cases  a  small  fraction 
of  the  steam  generated  will  run  a  steam  engine  or  steam  tur- 
bine, and  run  a  fan  capable  of  furnishing  all  the  draft  needed. 
In  this  manner  considerable  steam  is  available  for  other  pur- 
poses, and  great  economy  is  effected. 

CHIMNEY  DRAFT. 

The  principles  involved  are  not  obscure  or  complicated. 
The  total  pull,  or  suction,  which  a  chimney  can  produce,  as- 
suming it  to  be  filled  with  hot  air,  is  simply  due  to  the  ascen- 
sive force  of  the  hot  air  inside,  and  the  measure  of  this  is  the 
difference  of  weight  of  the  chimney  full  of  hot  gases  and  what 
it  would  be  if  filled  with  cold  air  of  the  temperature  outside. 

Illustration:  A  chimney  is  6  feet  square  inside  and  100  feet 
high,  uniform,  with  the  gases  inside  at  an  average  tempera - 
,ture  of  500°  F.,  and  specific  gravity  (air  =  1)  of  1.06.  The 

185 


186  METALLURGICAL  CALCULATIONS. 

air  outside  is  at  80°  F.  What  is  the  ascensive  force  of  the  hot 
gas  inside,  in  total  pounds,  in  ounces  per  square  inch  and 
inches  of  water  gauge  ? 

The  volume  of  the  space  in  the  chimney  —  chimney  volume  — 
is  100X6X6  =  3600  cubic  feet.  This  volume,  filled  with  air 
at  32°  F.,  would  weigh 

3600X1.293  =  4654.8  oz.  av.  =  290.9  pounds. 
And,  filled  with  gas  at  500°  F., 

491 
290.9X1.06X500_32  +  491  =  157.9  pounds. 

If  filled  with  outside  air,  at  80°  F.,  the  weight  would  be 

4Q1 


We,  therefore,  see  that  the  hot  gases  in  the  chimney,  weigh- 
ing 157.9  pounds,  displace  265.0  pounds  of  cold  air,  and  the 
tendency  of  the  former  to  rise  upwards  in  this  ocean  of  air 

must  be 

265.0—157.9  =  107.1  pounds. 

To  put  it  in  another  way,  if  a  piston  fitted  into  the  chimney 
at  the  bottom,  and  could  move  without  friction,  the  piston 
would  have  to  be  loaded  with  107.1  pounds  to  keep  it  from 
moving  up  the  chimney.  The  total  upward  pull  of  the  chimney 
is  therefore  107.1  pounds. 

Since  this  would  be  exerted  on  a  piston  6X6  =  36  square 
feet  in  area,  the  pull  or  suction  per  square  foot,  in  pounds,  is 
107.1-7-36  =  2.98  pounds,  and  in  ounces  per  square  inch. 

(2.98-5-144)  X  16  =  0.331  ounce  per  square  inch. 

If  the  pull  or  suction  is  measured  on  a  gauge,  as  by  water 
pressure,  the  pressure,  of  a  1-foot  column  of  water  at  ordi- 
nary temperatures  is  1000  ounces  per  square  foot,  or  1  inch  of 
water  is 

(  1  000  -T-  144)  -7-  12  =  0.597  ounces  per  square  inch. 
The  total  pull  of  the  chimney  is  therefore  equivalent  to 

'         =  0.57  inch  of  water  gauge. 

U  .  U4J7 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  187 

By  exactly  similar  methods  of  calculation  the  theoretical  total 
suction  of  a  chimney  of  any  given  height  and  temperature  of 
gases  inside  and  of  air  outside  may  be  obtained.  The  suction 
expressed  in  ounces  per  square  inch,  or  in  water  gauge  is,  of 
course,  independent  of  the  cross-sectional  area  of  the  chimney; 
it  depends  only  on  its  height  and  on  the  temperatures  inside 
and  outside. 

The  above  calculated  total  suction  (allowing  nothing  for 
friction,  etc.)  is  called  the  total  head  of  the  chimney,  and  is 
usually  expressed  in  terms  of  cold  air  (at  0°  C.)  instead  of  in 
water.  Cold  air  is  a  fluid,  and  water  is  772  times  as  heavy  as 
it;  therefore,  a  gauge  pressure  or  hydrostatic  head  of  0.57 
inch  of  water  is  the  same  as 

0.57X772  =  440  inches  of  air. 
=    36.5  feet  of  air. 

What  this  head  really  represents  is  clearly  seen  from  the 
above  calculation.  Its  value  is  to  be  obtained  directly  from 
the  height  of  the  chimney,  temperature  inside  and  out  and 
specific  gravity  of  the  chimney  gases  (air  =  1)  by  the  fol- 
lowing relations,  in  which 

h0  =  total  head  in  feet  of  air  at  32°  F. 

h0  =  total  head  in  meters  of  air  at  0°  C. 

t  =  temperature  in  the  chimney,  F°. 

t  =  temperature  in  the  chimney,  C°. 

t'  =  temperature  of  outside  air,  F°. 

t'  =  temperature  of  outside  air,  C°. 

D  =  Specific  gravity  of  chimney  gases,  air  =  1. 

H  =  Height  of  chimney  in  feet. 

H  =  Height  of  chimney  in  meters. 

coef  =  coefficient  of  gaseous  expansion,  F° 


coef  =  Coefficient  of  gaseous  expansion,  C°  =  -^=^ 


r(i— 

l°  Ql  +coef  (tf— 32)]  [1+coef  (t— 32) 

r(l— D  +  coef  (t— DtT| 
le~  (l  +  at')(l+crt)      J 


188  METALLURGICAL  CALCULATIONS. 

The  author  is  not  fond  of  using  formulas  whenever  their 
use  can  be  avoided.  The  above  formulas  express  in  the  sim- 
plest mathematical  form  the  principles  which  have  been  so 
far  explained  and  used  in  the  calculations,  but  it  is  strongly 
urged  that  the  formulas  be  kept  "  for  exhibition  purposes  only," 
and  that  when  any  specific  case  is  to  be  worked  it  be  attacked 
from  the  standpoint  of  the  principles  involved,  as  explained  in 
the  case  worked.  In  other  words,  if  one  understands  properly 
and  thoroughly  the  basic  principles,  he  has  no  need  of  the 
formula;  if  one  does  not  understand  the  principles,  the  formula 
had  better  be  kept  forever  in  "  innocuous  desuetude." 

The  total  head,  obtained  as  above,  is  the  theoretical  head. 
It  is  like  the  pressure  on  the  piston  of  a  locomotive  —  the  total 
available  force  for  all  purposes.  Just  as  the  pressure  on  the 
locomotive  piston  is  used  up  in  friction  in  the  engine  and  in 
moving  the  engine  itself,  and  the  residue  is  the  available  pull 
on  the  draw-bar  which  moves  the  train,  so  the  total  head  of 
the  chimney  is  partly  used  up  in  friction  in  the  chimney  itself, 
partly  in  giving  velocity  to  the  gases  as  they  pass  out  of  the 
chimney,  and  the  residue  is  the  available  head  which  draws  or 
pulls  the  gases  through  fire-grates,  furnaces  and  flues  up  to 
the  base  of  the  chimney.  If  the  chimney  could  be  momen- 
tarily completely  closed  at  the  bottom,  except  for  the  gauge 
opening,  and  the  gas  inside  be  brought  to  rest,  the  gauge  would 
show  the  total  head  ;  as  soon  as  dampers  are  opened  connecting 
the  flues,  gas  moves  up  the  chimney,  and  the  gauge  pressure 
is  lessened  by  the  head  required  to  move  the  gases  and  that 
absorbed  in  the  friction  in  the  chimney. 

Head  Represented  in  Velocity  of  Issuing  Gases.  —  This  item 
always  exists  .when  the  chimney  is  working,  and  depends  only 
on  the  velocity  of  the  gases  as  they  escape  and  their  tempera- 
ture. The  hydraulic  head  necessary  to  give  any  fluid  a  velocity 
V  is  simply  the  same  as  the  height  which  a  falling  body  must 
fall  in  order  to  acquire  that  same  velocity;  i.e.'. 


in  which  expression  g  is  the  constant  acceleration  of  gravity, 
9.8  meters  or  32.2  feet,  and  the  velocity  is  in  meters  or  feet 
per  second.  If  we  know,  therefore,  the  velocity  of  the  gases 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  189 

issuing  from  the  chimney,  or  can  calculate  or  assume  it,  we 
can  get  h.  In  practice  the  velocity  does  not  vary  within  very 
wide  limits.  In  small  house  chimneys  it  may  not  exceed  3  feet 
per  second,  in  boiler  chimneys  6  to  12  feet  per  second,  in  fur- 
nace chimneys  12  to  20  feet  per  second.  The  temperatures  of 
these  issuing  gases  is,  moreover 

C°  F° 

In  small  chimneys  ........  100  to    200  200  —  350 

In  boiler  chimneys  ........  100  to    300  200°  —  550 

In  furnace  chimneys  ......  300  to  1000  550°—  18QO° 

If  the  chimney  in  question  has,  therefore,  a  known  velocity 
of  exit  of  its  gases,  h  can  be  calculated;  but  it  must  not  be 
forgotten  that  h  will  be  in  terms  of  the  kind  of  gases  which 
is  escaping;  i.e.,  of  hot  gas,  and  to  subtract  it  from  or  com- 
pare it  with  h0  we  must  reduce  it  to  its  equivalent  head  in 
terms  of  cold  gas.  This  is  merely  a  matter  of  taking  into 
account  the  specific  gravities  or  relative  densities  of  hot  gas 
and  cold  air,  which  are  inversely  proportional  to  their  absolute 
temperatures;  that  is,  if  D  represents  the  relative  density  of 
air  and  chimney  gases  at  the  same  temperature: 

V,     vel.   —  V,  V 

h«      hxi+«t" 

Illustration:  Assuming  the  actual  velocity  of  the  gases  is- 
suing from  a  furnace  chimney  to  be  15  feet  per  second,  and 
their  temperature  500°  F.,  density  1.06  (air  =  1),  what  will 
be  the  head  represented  by  the  velocity  of  these  gases  in  terms 
of  cold  air  at  32°  F.  ? 

The  head  represented,  in  terms  of  hot  gases  at  500°  F.,  is 


In  terms  of  air  at  500°  F.  is 

3.5X1.06  =  3.71  feet. 
And  in  terms  of  air  at  32°  F., 

4Q1 


190  METALLURGICAL  CALCULATIONS. 

Out  of  the  total  head  which  this  chimney  produces  (say  36.5 
feet),  1.9  feet  is  represented  by  the  velocity  of  the  issuing 
gases,  or  5.2  per  cent,  of  the  whole,  leaving  34.6  feet  to  repre- 
sent loss  by  friction  in  the  chimney  and  the  available  head. 
We  will  proceed  to  discuss  the  loss  of  head  due  to  friction  in 
the  chimney. 

Head  Lost  in  Friction  in  the  Chimney. — This  varies  with 
the  smoothness  or  roughness  of  the  walls,  and  has  been  deter- 
mined experimentally  for  air  moving  with  different  velocities. 
'  The  manner  of  expressing  the  friction  loss  is,  to  put  it  as  a 
function  of  the  head  necessary  to  give  the  gases  their  actual 
velocity,  assuming  there  were  no  friction.  Thus,  supposing  as 
in  the  preceding  paragraph,  the  actual  velocity  of  the  hot 
gases  is  15  feet  per  second,  and  the  head  (in  terms  of  cold 
air)  necessary  to  give  that  velocity,  not  considering  friction, 
is  1.9  feet,  then  the  head  lost  in  friction  in  getting  up  this 
velocity  will  be 

h  friction  =  i.gx-^K. 
d 

That  is,  it  will  be  proportional  to  H,  the  height  of  the  chimney, 
inversely  as  d,  the  diameter  or  side,  if  square,  and  to  a  co- 
efficient K,  determined  by  experiment.  The  latter  varies,  ac- 
cording to  Grashof  s  experiments,  between  0.05  for  a  smooth 
interior  to  0.12  for  a  rough  one,  and  averages  0.08. 

Illustrations'.  Assuming  the  height  of  the  chimney,  100  feet, 
its  section  to  be  6  feet  square,  the  coefficient  of  friction  K  = 
0.08,  and  the  head  represented  by  the  net  velocity  of  the  hot 
gases  in  the  chimney  to  be  1.9  feet  of  cold  air,  what  is  the 
head  lost  in  friction  in  the  chimney? 

The  ratio  of  height  to  side  is  100-7-6  =  16.67,  which  mul- 
tiplied by  K  gives  1.33  as  the  value  of  the  function  containing 
these  three  terms.  This  means  that  1.33  times  as  much  head 
has  been  lost  in  friction  as  is  represented  by  the  net  actual 
velocity  of  the  gases  as  they  pass  up  the  chimney.  Therefore, 

h  friction  =  1.9X1.33  =  2.5  feet  cold  air. 

Another  way  of  looking  at  this,  which  is  sometimes  useful 
in  considering  the  height  of  a  chimney,  is  to  say 

h  £riction  =  0.025  H. 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  191 

Or,  that  in  this  case,  the  head  lost  in  friction  amounts  numer- 
ically to  one-fortieth  the  height  of  the  chimney. 

If  we  subtract  the  head  lost  in  friction  plus  that  represented 
in  the  net  velocity  of  the  gases,  from  the  total  gross  head,  the 
residue  is  that  available  for  doing  work  external  to  the  chimney. 
In  the  specific  case  of  the  preceding  illustrations  we  have 

h0  =  total  head  =  36.5  feet  =  100  per  cent, 

h  velocity   =  velocity  hea'd  =    1.9    "     =  ~~5 

k  friction    =  friction  in  chimney  =    2.5    "     =      7 

h  available  =  available  head  =32.1    "     =    88 

Available  Head  of  a  Chimney. — This  is  the  part  of  the  total 
head  which  remains  after  subtracting  the  head  lost  in  friction 
in  the  chimney  and  that  represented  by  the  velocity  of  the 
issuing  gases.  In  the  specific  cases  considered  in  the  above 
illustrations,  the  net  available  head  amounted  to  88  per  cent, 
of  the  whole  theoretical  head.  If  we  assume  limiting  con- 
ditions as  found  in  practice,  we  can  find  the  limiting  values  of 
this  proportion.  Calling  the  cases  I  and  II,  those  with  mini- 
mum and  maximum  absorption  of  head  in  the  chimney  itself, 
we  have 

Case  I.     Case  II. 

Temperature  of  issuing  gases 100°  C.     1000°  C. 

Velocity  of  issuing  gases  per  second 1  meter    7  meters 

Ratio  H  to  d 10  50 

Coefficient  K 0.05          0.12 

Specific  gravity  of  gases  (air  =  1) 1.00          1.06 

Head  as  velocity  of  gases  (meters  of  air).  .0.04m.    0.56m. 

Head  as  velocity  of  gases  (feet  of  air) 0.13          1.87 

Head  absorbed  in  friction   (meters  of  air) .  .0.02          3.36 

Head  absorbed  in  friction  (feet  of  air) 0.07          11.2 

Head  used  up  in  chimney  (meters) 0.06  to     3.92 

Head  used  up  in  chimney  (feet) .  0.20  to     13.0 

Water  gauge  pressure  thus  lost,  m.m 0.1  to       5.0 

Water  gauge  pressure  thus  lost,  inches 0.003  to  0.2 

The  available  head,  will,  therefore,  be  the  theoretical  total 
head  minus  a  loss  in  the  chimney  itself,  which  may  amount 
to  a  maximum  of  3.9  meters  or  13  feet,  representing  an  ab- 
sorption of  water  gauge  pressure  up  to  5  millimeters,  or  0.2 


192  METALLURGICAL  CALCULATIONS. 

inch   at   a  maximum.     Under   ordinary  conditions  half  these 
quantities  would  be  a  rather  high  chimney  loss. 

In  most  conditions  which  confront  the  metallurgist,  the 
question  is  to  determine  how  high  a  chimney  should  be  built 
in  order  to  supply  a  certain  available  draft  determined  by 
practice  to  be  necessary.  For  instance,  to  burn  a  certain 
amount  of  coal  per  hour  on  any  grate  requires  a  certain  amount 
of  draft.  This  amount  is  increased  if  the  draft  is  increased, 
and  vice  versa.  In  boilers,  18  pounds  of  coal  burned  per  square 
foot  of  grate  surface  per  hour  is  highly  economical  practice, 
and  requires  a  draft  of  0.4  inch  to  0.8  inch  of  water  gauge, 
according  to  the  kind  of  coal  'burned.  In  furnaces  where  the 
amount  of  coal  burned  is  greater  per  hour  there  will  be  usually 
a  correspondingly  greater  temperature  in  the  chimney.  To 
calculate  the  height  of  chimney  required  it  is  necessary  to  assume 
only  the  temperature  in  the  chimney  >  the  available  draft  re- 
quired and  an  average  chimney  loss. 

Problem  22. 

It  is  desired  to  design  a  chimney  for  a  puddling  furnace,  the 
grate  of  which  is  4  feet  by  6  feet,  and  which  shall  burn  30  pounds 
of  bituminous  coal  per  hour  per  square  foot  of  grate  surface. 
Temperature  of  gases  entering  the  chimney  1200°  C.,  at  the 
top  probably  1000°  C.  Specific  gravity  of  gases  1.03  (air  =  1). 
Draft  required  0.6  inch  of  water  gauge.  Outside  temperature 
30°  C. 

Solution:  We  can  assume  that  since  the  gases  will  be  at 
an  average  temperature  of  1100°  C.  in  the  chimney,  their  ve- 
locity will  be  high,  and  that  at  least  0.1  inch  of  water  gauge 
pressure  will  be  absorbed  by  the  chimney  itself.  This  makes 
a  total  requirement  of  0.7  inches  of  water  for  total  head,  or 

h0  =  0.7X772^12  =  45  feet  of  cold  air. 
Or  an  unbalanced  pressure  or  ascensive  force  of 

45X1.293'-*-  16  =  3.64  pounds  per  square  foot. 

Considering  the  air  outside  the  chimney,  its  weight  at  30°  C. 
is  equal,  per  cubic  foot, 


1.293  X  -16  =  0.073  pounds. 

oUo 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  193 

The  gases  inside  the  chimney  weigh,  per  cubic  foot, 

273 
1.293Xl.03Xn()Q  +  273^16  =  0.0166  pounds. 

The  height  of  the  chimney  being  called  H  and  its  cross- 
section  S,  the  volume  is  HxS,  and  the  weight  of  hot  gas  inside 
it  is 

(HXS)X  0.0166  pounds. 

And  of  an  equal  volume  of  cold  air  outside 

(H  X  S)  X  0.073  pounds, 
giving  a  total  ascensive  force  of 

(H  X  S)  X  0.0564  pounds. 
But  there  is  needed  a  total  ascensive  force  of 

SX3.64  pounds, 

in  order  to  give  the  pull  of  3.64  pounds  per  square  foot,  and, 
therefore,  of  necessity, 

HXSX  0.0564  =  SX3.64, 
from  which 


Concerning  the  cross-section  of  this  chimney,  it  would  not 
be  safe  to  make  it  less  in  diameter  than  one-fiftieth  of  its  height, 
because  of  lack  of  stability;  in  fact,  one  -twenty-fifth  would 
be  better  practice.  This  consideration  would  make  its  in- 
ternal diameter  2  ft.  7  inches,  area  5.2  square  feet.  Another 
way  of  arriving  at  a  diameter  is  to  calculate  the  volume  of  the 
hot  gases  which  must  pass  up  the  chimney  and  assume  for  them 
some  maximum  velocity  in  the  chimney,  such  as,  let  us  say, 
6  meters  (20  feet)  per  second,  and  so  get  the  minimum  area 
necessary  for  filling  this  condition  as  follows: 

Coal  burnt  per  hour  4  X  6  X  30  =          720  pounds. 

Air  theoretically  necessary,  assuming  aver- 

age bituminous  coal  (see  Prob.  1)  =  123 

X  720  =    88,560  cubic  feet. 


194  METALLURGICAL  CALCULATIONS. 

Products  of  combustion  at  standard  condi- 
tions =  129X720  =    92,880  cubic  teet 
Volume  chimney  gases  at  1100°  C.  = 

. 467,100      -. 

Volume  per  second  =         130 

Area  of  chimney,  if  maximum  velocity  is  20 

feet  per  second  =         6 . 5  sq.  feet 

Diameter,  if  round  =  2  ft.  10  in. 

This  chimney  would  do  its  work  better,  and  there  would  be 
much  less  loss  in  friction,  if  the  internal  diameter  were  made  25 
per  cent,  greater  than  the  above  calculated  minimum,  say,  there- 
fore, 3  feet  6  inches,  making  the  area  nearly  50  per  cent,  greater 
and  cutting  down  the  velocity  in  the  chimney  to  13.5  feet  per 
second. 

Problem  23. 

In  the  case  of  the  puddling  furnace  of  Problem  22,  assume 
that  the  hot  gases,  instead  of  going  directly  into  the  chimney, 
are  passed  through  the  flues  of  a  boiler  placed  above  the  fur- 
nace, and  thence  pass  into  the  chimney  at  a  point  15  feet 
higher  than  before.  Assume  chimney  3  feet  6  inches  internal 
diameter,  64.5  feet  high  above  the  furnace  flue,  and  that  the 
gases  now  passing  into  it  15  feet  higher  up  are  at  350°  C.,  and 
cool  to  250°  C.  at  the  top  of  the  chimney.  The  boiler  flues  in- 
troduce additional  frictional  resistance  equal  to  0.1  inch  of 
water.  The  boiler  raises  steam  at  a  net  efficiency  of  45  per 
cent.,  the  steam  engine  utilizes  the  steam  at  a  mechanical 
efficiency  of  20  per  cent.,  and  a  centrifugal  fan  supplies  the 
forced  draft  needed  at  a  mechanical  efficiency  of  25  per  cent. 

Required:  (1)  The  total  head  of  the  chimney,  when  the 
furnace  discharged  directly  into  it,  and  the  average  tempera- 
ture of  the  gases  in  it  was  1100°  C.,  and  specific  gravity  1.03 
(air  =  1). 

(2)  The  head  absorbed  as  velocity  of  the  outgoing  gases, 
their  temperature  being  1000°  C. 

(3)  The  head  lost  in  friction  in  the  chimney,  in  this  case. 

(4)  The  head  which  was  available  to  run  the  puddling  furnace. 

(5)  The  total  head  of  the  chimney  with  the  gases  entering 
15  feet  above  former  flue,  and  average  temperature  300°  C. 


CHIMNEY  DRAFT  AND  FORCED  DRAFT  195 

(6)  The  head  absorbed  in  this  case  as  velocity  of  outgoing 
gases,  their  temperature  being  250°  C. 

(7)  The  head  lost  in  friction  in  the  chimney  in  this  case. 

(8)  The  available  head  to  draw  gases  into  the  chimney. 

(9)  The  deficit  of  head  which  must  be  made  up  by  forced 
blast  under  the  grate  of  puddling  furnace. 

(10)  The  horse-power  absorbed  by  the  fan  which  furnishes 
this  blast. 

(11)  The    horse-power   furnished   by    the    engine    using   the 
steam  from  the  boiler. 

(12)  The  excess  of  power  which  is  thus  saved  and  available 
for  other  purposes. 

Solution  : 
(1)  Volume  of  gases  in  chimney: 

64.5X3.5X3.5X0.7854  =  620.5  cu.  ft. 
Weight  at  32°  F.  (0°  C.): 

620.5  X  (1.293  4-  16)  X  1.03  =  51.65  Ibs. 
Weight  if  temperature  is  1100°  C.: 
070 


Weight  of  equal  volume  of  air  outside  at  30°  C.: 

070 

620.  5X  (1.293-;-  16)  Xon  ,  *      =  45.18  Ibs 

oU  -T  Z<  o 

Difference  of  weight  =  ascensive  force, 

45.18—10.27  =  34.91  Ibs. 
Ascensive  force  per  square  foot, 

34.91  -v-  9.62  =  3.63  Ibs. 
Total  head  in  terms  of  cold  air  at  0°  C., 

3.63  -T-  (1.293  H-  16)  =  44.9ft.  (1) 

In  terms  of  water  gauge  pressure, 

44.9X12-772  =  0.685  ins.  (1) 


196  METALLURGICAL  CALCULATIONS. 

(2)    Volume  of  gases  per  hour  at  0°  C., 

(Prob.  22)  =  92,880  cu.  ft. 

Volume  at  1000°  C., 

=  92,880  X10°°1:273  =  433,100  cu.  ft. 


Velocity  per  second, 

433,  100  -h  (3600)  -9,  62  =  12.50ft. 

Head  necessary  to  give  this  velocity,  in  terms  of  hot  gases, 
at  1000°  =  (12.50)2-^-64.3  (2g)  =  2.43ft. 

In  terms  of  gases  at  0°  C., 


In  terms  of  air  at  0°  C., 

0.52X1.03  =  0.55ft.  (2) 

In  terms  of  water  gauge  pressure, 

0.55  X  12  -J-  772  =  0.008  in.  (2) 

(3)    Assuming  K,  the  coefficient  of  friction,  0.08,  then 


h    friction  =  __      H 

2g     273  +t      d 

This  is  only  an  abbreviated  form  of  the  operations  done 
under  (2),  adding  the  terms  which  account  for  the  height, 
diameter  and  friction.  Now,  the  velocity  per  second: 

i  inn  -4-  97*} 
V  =  92,880  x"Jl       -H  3600-^9.62  =  13.5ft. 


Head  necessary  to  give  this  velocity  in  terms  of  air  at  0°, 

070 
a3.5)'H-64.3Xg?gTn55Xl.03  =  0.58  ft. 

Proportion  of  this  velocity  head  lost  in  friction  = 

£  K  -  ^XO.08  -  1.47     ' 
a  3.5 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  197 

/ 

Head  lost  in  friction  in  chimney, 

0.58X1.47  =  0.85ft.  (3) 

In  terms  of  water  gauge  pressure, 

0.85X12*772  =  0.013  in.  (3) 

(4)  Cold  Air.  Water  Gauge. 
Total  head  .......................  44.90  feet  0.685  inch 

Absorbed  in  velocity  of  gases.  ......   0.55    "  0.008     " 

Absorbed  in  friction  in  chimney  ....   0.85    "  0.013     " 

Available  for  the  furnace  ...........  43.50    "  "O664     "     (4) 

(5)  Volume  of  chimney  gases, 

(64.5—  15)  X  3.5X3.5X0.7854  =  475.7  cu.  ft. 

Weight  at  300°  C.,  specific  gravity  1.03  (air  =  1). 

07Q 

475  .  7  X  (1  .  293  *  16)  X  1  03  X  273  +  300  =  18.86  Ibs. 

Weight  of  equal  volume  of  outside  air  at  30°  C., 

273 
475.  7X  (1.293*  16)  X273™30  =  34.68  Ibs. 

Ascensive  force  of  air  per  square  foot, 

(34.68—  18.86)  -9.62  =  1.64  Ibs. 
Total  head  in  terms  of  cold  air, 

1.64*  (1.293-5-16)  =  20.3  ft.  (5) 

In  terms  of  water  gauge  pressure, 

20.3X12*772  =  0.32  in.  (5) 

(6)  Velocity  of  issuing  gases,  per  second,  at  250°  C., 

OCQl    070 

92,880  X      0I"'    -5-3,600*9.62  =  5.14ft 


Head  as  velocity  in  terms  of  cold  air  at  0°  C., 


(5.  14)2*  64  3X^X1.03  =  0.22ft, 

In  terms  of  water  gauge  pressure  =  0.003  in. 
(7)    Average  velocity  of  gases  in  chimney  at  300°  C., 

onn  I  970 

92,880  *  3,600  X-01T      *9.62.=  5.63ft. 

AIO 


198 


METALLURGICAL  CALCULATIONS. 

\ 

Head  lost  in  friction  in  terms  of  cold  air, 

27S  4Q   5 

(5.63)2-^54.3X3QO|273Xl.03X-^yX0.08  =  0.27ft.      (7) 

In  terms  of  water  gauge  pressure  =  0.004  in.  (7) 

(8)  Cold  Air.  Water  Gauge. 

Total  head  ....................  .  .  .20.30  feet        0.320  inch 

Absorbed  in  velocity  of  gases  .......   0.22    "          0.003     " 

Absorbed  in  friction  in  chimney  ----   0.27    "          0.004     " 

Available  to  draw  gases  in,  ........  19.81    "          0.313     "     (8) 

(9) 

Available  head  needed  for  puddling 

(4)...  .........................  43.50  "  0.664  " 

Available  head  needed  for  boiler.  .  .  .   6.43  "  0.100  " 

Total  head  needed  for  both  ........  49.93  "  -    0.764  " 

Available  head  from  chimney  (8).  .  .19:81  "  0.313  " 

Deficit,  to  be  supplied  by  blast  .....  30.12  "  0.451  " 

(10)  The  0.451  inches  of  water  gauge  equals 

(0.451  -5-1  2)  X  62.5  =  2.35  Ibs.  per  sq.  ft. 

The  volume  of  air  to  be  supplied  is,  at  30°  C., 
88,560  (Prob.  22)^60X^|^^  =  1,638  cu.  ft.  per  min. 


Net  work  dpne  by  the  fan, 

1,638X2.35  =  3,850  ft.  Ibs.  per  min 

Gross  power  needed  by  the  fan, 

3,850-7-0.25  (efficiency)  =  15,400  ft.  Ib.  per  min. 

Horse-power  needed  to  drive  the  fan, 

15,400-^-33,000  =  0.47  H.  P.  (10) 

(11)  The  boiler  receives  the  gases  at  1200°  C.  and  discharges 
them  at  350°,  and  92.880  cubic  feet  of  gases  (measured  at 
standard  conditions)  pass  through  per  hour.  The  composition 
of  these  gases  is  not  given,  but  from  the  specific  gravity  we 
might  conclude  that  they  contain  on  an  average  10  per  cent,  of 
carbon  dioxide,  since  if  they  contained  the  maximum  amount 


CHIMNEY  DRAFT  AND  FORCED  DRAFT.  199 

of  that  gas  (about  20  per  cent.)  their  specific  gravity  would  be 
1.06  (air=l).     Assuming  them,  therefore,  to  contain 

CO2 10  per  cent. 

H2O . 10      "  " 

CO,  N2,  O2 80 

their  heat  capacity  per  degree  per  cubic  foot  would  be,  between 
350°  and  1200°. 

CO2  0.1X[0.37  +0.00022  (350  +  1200)]  =  0.0711  oz.  cal. " 

H2O  0. IX [0.34   +0.00015  (350  +  1200)]  =  0.0573     " 

CO,  N2,  O2  0.8X[0.303  +  0.000027 (350  + 1200)]  =  0.2759    " 


Sum  =  0.4043 
Heat  given  up  per  cubic  foot, 

0.4043  X  (1200— 350)  =  343.7  oz.  cal. 

Heat  given  up  by  gases  per  hour  to  boiler, 
92,880X343.7  =  31,922,850  oz.  cal. 
1,995,000  Ib.  cal. 

Heat  in  the  steam  produced  per  hour, 

1,995,000X0.45  (efficiency)  =  897,750  Ib.  cal. 

Heat  equivalent  of  mechanical  energy  of  steam  engine  per 
hour, 

897,750X0.20  (efficiency)  =  179,950  Ib.  cal. 
Heat  equivalent  of  1  hp  hour  =        635  kg.  cal. 

=     1,400  Ib.  cal. 

Horse-power  generated  by  the  engine, 

179,950-5-1,400  =   128  H.  P.  (11) 

(12)  Net  available  power  after  supplying  fan, 

128—0.5  =  127.5  H.  P.  (12) 


CHAPTER  VIII. 
CONDUCTION  AND  RADIATION  OF  HEAT. 

These  two  factors  are  of  the  greatest  practical  importance 
to  the  metallurgist,  yet  they  are  also  the  subjects  of  all  others 
upon  which  the  practical  metallurgist  is  usually  the  most 
poorly  informed.  It  often  happens  that  a  furnace  is  built  of 
twice  the  capacity  of  its  predecessor  or  of  its  neighbors,  and 
the  manager  unexpectedly  and  most  agreeably  discovers  that 
it  keeps  up  a  more  uniform  heat  and  requires  considerably  less 
than  twice  the  amount  of  fuel  to  keep  it  running.  Two  reasons 
were  operative;  first,  the  walls  were  made  thicker  to  support 
the  heavier  structure,  but  that  made  them  also  poorer  conduc- 
tors of  heat;  second,  the  capacity  increased  as  the  cube  of  a 
linear  dimension,  while  the  radiating  surface  increased  as  its 
square,  so  that  radiation  was,  therefore,  less  than  twice  the 
primary  amount  from  the  furnace  of  double  capacity.  Many 
practical  metallurgists  have  learned  the  practical  results,  but 
are  entirely  ignorant  of  the  principles  upon  which  the  results 
are  obtained.  While  it  is  well  to  be  successful,  it  is  better  to 
be  intelligently  so. 

PRINCIPLES  OF  HEAT  CONDUCTION. 

It  is  well  known  that  the  ability  of  metals  to  conduct  heat 
is  very  nearly  the  same  as  their  ability  to  conduct  electricity. 
The  order  of  metals  in  these  two  series  is  almost  identical. 
Further,  the  specific  conductance  for  heat  is  closely  analogous 
to  specific  conductance  for  electricity.  Just  as  we  say  that  the 
electrical  resistance  of  a  conductor  is  proportional  to  its  length 
and  inversely  as  its  cross-section,  so  we  can  make  the  same 
statement  concerning  heat  resistance.  Just  as  we  can  add 
electrical  resistances  when  the  bodies  are  in  series,  and  must 
add  electrical  conductances  when  they  are  in  parallel,  so  we 

200 


CONDUCTION  AND  RADIATION  OF  HEAT.  201 

can  add  heat  resistances  when  the  bodies  are  in  line,  and  must 
add  heat  conductances  when  they  are  hooked  up  together  in 
parallel. 

The  unit  of  electrical  resistance  is  an  ohm,  and  a  substance 
has  unit  resistivity  when  a  cube  of  it,  1  centimeter  on  a  side, 
has  that  resistance.  In  such  a  case  a  drop  of  potential  of  1 
volt  from  one  side  to  the  opposite  one  sends  through  the  cube 
1  coulomb  of  electricity  per  second.  Since  conductivity  is 
merely  the  reciprocal  quality,  actually  and  mathematically,  to 
resistivity,  the  unit  of  conductivity  is  defined  in  exactly  the 
same  way,  and  the  conductivity  of  any  substance  may  be  ex- 
pressed as  so  many  reciprocal  ohms. 

If  we  read  in  the  preceding  paragraph,  thermal  resistance 
for  electrical  resistance,  degrees  temperature  for  volts,  and 
gram  calories  for  coulombs,  we  have  defined  the  corresponding 
unit  of  thermal  resistivity.  A  substance  has  unit  capacity 
for  conducting  heat  when  a  cube  1  centimeter  on  a  side  trans- 
mits 1  gram  calorie  of  heat  per  second,  with  a  drop  of  tem- 
perature from  one  surface  to  the  other  of  1°  C.  Many  investi- 
gators have  adopted  slightly  different  units,  such  as  1-kilo- 
gram Calorie  per  cubic  meter  per  hour  per  degree  difference; 
but  such  are  only  simple  multiples  or  fractions  of  the  centi- 
meter-gram-second unit  above  denned. 

Illustration'.  If  the  thermal  conductivity  of  copper  is  0.92 
units,  what  would  be  the  amount  of  heat  passing  per  hour 
through  a  sheet  1  millimeter  thick  and  1  meter  square,  with 
a  constant  difference  of  1°  C.  between  the  two  sides? 

Solution:  The  0.92  units  means  that  a  column  of  copper  1 
centimeter  long  and  1  square  centimeter  in  cross-section,  hav- 
ing a  difference  of  temperature  at  its  two  surfaces  of  1°  C., 
would  allow  0.92  gram-calories  of  heat  to  flow  through  it  per 
second.  For  a  sheet  one-tenth  as  thick,  10,000  times  the  area 
and  during  1  hour,  the  heat  passing  per  1°  C.  difference  will  be 

0.92XyX10^°0  X360Q  =  331,200,000  cal. 

=  331, 200  Cal. 

Such  calculations  as  the  above  are,  of  course,  very  useful  for 
comparing  different  thicknesses  of  plates  of  different  material, 
as  to  their  relative  heat-carrying  capacity.  With  some  slight 


202  METALLURGICAL  CALCULATIONS. 

modifications  they  are  applicable  to  the  actual  conveyance  of 
heat  through  such  walls,  from  a  fluid  on  one  side  to  a  fluid  on 
the  other.  This  introduces  an  idea  analogous  to  transfer  or 
contact  resistance  in  electrolytic  conduction.  Thus,  if  we  call 
R  the  thermal  specific  resistance  (resistivity)  in  C.  G.  S.  units, 
of  the  material  of  a  partition  having  a  thickness  of  d  centi- 
meters, and  an  area  of  S  sqtiare  centimeters,  then  the  thermal 
resistance  of  the  body  of  the  partition  will  be 

R  X  d 


and  its  thermal  conductance 

-£-     =k 
R  X  d 

Besides  this  resistance  in  the  body  of  the  material  there  is 
transfer  resistance  at  its  two  sides,  or  at  its  inner  and  outer 
surfaces,  which  depends  on  the  materials  which  communicate 
heat  and  receive  heat,  and  their  velocity.  These  resistances 
may  be  expressed  as  so  many  units  per  square  centimeter,  the 
unit  being  of  exactly  the  same  nature  as  R,  except  that  it 
connotes  no  thickness  but  merely  a  surface  effect.  Calling  R1 
the  specific  resistance  of  transfer  from  the  inner  fluid  to  the 
inner  surface  of  the  material,  and  R2  the  outside  specific  trans- 
fer resistance  to  the  fluid  outside,  we  can  consider  all  three 
resistances  as  being  in  series,  and  the  total  thermal  resistance 
to  transfer  from  the  inside  fluid  to  the  outside  fluid  to  be 

R  X  d    R1     R2 


or  the  thermal  conductance  of  the  system 


(RX<*)-fR' 

The  following  table  gives  the  thermal  resistivity  of  various 
materials,  in  C.  G.  S.  gram-Calorie  units,  also  the  thermal 
conductivity  in  units  which  are  reciprocals  of  the  resistance 
units: 


CONDUCTION  AND  RADIATION  OF  HEAT. 


203 


Material. 


Conductivity 

(in  reciprocal 

resistance  units . ) 


Silver  (at  0°) 1.10 

Copper  (0°— 30°). . '. 0.92 

Copper,  commercial 0.82 

Copper,  phosphorized 0. 72 

Magnesium 0 . 38 

Aluminium  (0°) 0.34 

Aluminium  (100°) 0.36 

Zinc  (15°) 0.30 

Brass,  yellow  (0°) 0.20 

Brass,  yellow  (100°) 0. 25 

Brass,  red  (0°). 0.25 

Brass,  red  (100°) 0.28 

Cadmium  (0°). , 0.20 

Tin  (15°) 0.15 

Iron,  wrought  (0°), 0.21 

Iron,  wrought  (100°). 0. 16 

Iron,  wrought  (200°). C.  14 

Iron,  steel,  soft 0.11 

Iron,  steel,  hard 0.06 

German  silver  (0°) 0.07 

German  silver  (100°) 0.09 

Lead  (0°) 0  084 

Lead  (100°) 0.076 

Antimony  (0°) ".  .  .  .  0.044 

Mercury  (0°)., 0.015 

Mercury  (50°) 0.019 

Mercury  (100°) 0.024 

Bismuth  (0°) 0.018 

Wood's  alloy  (7°) 0.032 

Allov,  1  Sn.  99  Bi.  .  0.008 


Resistivity 
(inC.G  S. 

gram-cal.  units.} 

0.91 


09 
22 
39 
63 

2  94 
2.75 
33 
00 
00 
00 
3.57 
5.00 
6.67 
4.76 
6.25 
7.14 
9.01 
16.67 
14.28 
11.11 
11.90 
13.16 
22.73 
66.67 
52.63 
41.67 
55.55 
31.25 
125.00 


The  last  alloy,  bismuth,  with  only  1  per  cent,  of  tin,  is -re- 
markable for  its  extremely  low  heat  conductivity,  less  than  one 
hundredth  as  good  as  copper.  It  has  also  the  lowest  electrical 
conductivity  of  any  alloy,  about  one  two  hundredth  that  of 
copper.  These  peculiar  properties  ought  to  make  it  of  use  for 
some  particular  purposes. 

The  above  data  enables  one  to  calculate  the  rate  at  which  heat 


204  METALLURGICAL  CALCULATIONS. 

will  pass  through  a  metallic  partition  or  wall  when  the  tempera- 
ture of  its  two  surfaces  is  known.  This  is  a  very  useful  calcu- 
lation, for  in  many  cases  the  temperature  inside  a  partition  or 
pipe  is  known,  or  can  be  easily  determined,  and  when  the  tem- 
perature of  the-  outside  surface  is  determined,  the  calculation 
can  be  made.  The  temperature  of  the  outside  of  a  partition  or 
pipe  can  be  found  by  several  methods:  one  is  to  lay  a  very  flat 
bulb  thermometer  (made  especially  for  this  purpose)  against 
it;  another  is  to  put  the  junction  of  a  thermocouple  against  it, 
covering  the  couple  with  a  little  putty  or  clay;  another  is  to 
take  small  pieces  of  metals,  or  alloys  of  known  melting  points, 
and  see  which  melts  against  the  hot  metal.  The  only  uncer- 
tainty then  in  the  calculation  is  the  question  as  to  what  differ- 
ence in  the  conductivity  may  be  caused  by  the  higher  tempera- 
ture, The  conductivities  in  most  cases  decrease  as  tempera- 
ture rises,  but  in  others  increase.  There  is  here  a  large  field 
for  metallurgical  experiment,  in  determining  the  heat  con- 
ductivities of  metals,  alloys  and  fire-resisting  materials  at  high 
temperature. 

Illustration'.  An  iron  pipe,  6  centimeters  in  diameter  out- 
side and  5  centimeters  inside,  is  filled  with  water  at  10°  C.,  and 
surrounded  by  hot  gases  at  198°  C.  From  the  rise  in  tempera- 
ture of  the  water  it  is  known  that  for  each  square  centimeter  of 
heating  surface  0.084-gram  calories  of  heat  pass  per  second. 
Assuming  the  thermal  conductivity  of  the  iron  (k)  =  0.14, 
what  is  the  difference  of  temperature  of  the  two  surfaces,  inside 
and  outside,  of  the  pipe?- 

Solution:  The  thickness  of  the  walls  is  (6—5)^2  =  0.5 
centimeter.  If  the  walls  were  1  centimeter  thick,  1°  difference 
would  transmit  0.14  calorie;  but  being  only  half  that  thick,  1° 
difference  would  transmit  0.28-gram  calorie.  The  actual  dif- 
ference of  temperature  of  the  two  surfaces  must  then  be 


Of  course,  such  calculations  can  be  turned  around,  and  if  the 
temperature  of  the  inside  and  outside  surfaces  is  known,  the 
heat  being  transmitted  can  be  calculated;  or  if  the  temperature 
of  these  two  surfaces  is  known  and  the  heat  being  transmitted 
is  measured,  the  thermal  conductivity  of  the  partition  can  be 


CONDUCTION  AND  RADIATION  OF  HEAT.  205 

reckoned;  or,  again,  if  the  temperature  of  the  two  surfaces  is 
known,  and  also  the  thermal  conductivity  of  the  partition,  and 
the  temperature  of  either  fluid  on  either  side,  the  thermal  re- 
sistance of  the  transfer  from  either  of  the  fluids  to  the  sur- 
face of  the  partition  can  be  calculated. 

Illustration',  In  the  previous  illustration  the  temperature  of 
the  hot  gases  was  198°,  while  that  of  the  pipe  in  contact  with 
them  was  practically  10°.  What  was  the  thermal  resistivity  of 
the  transfer  from  gas  to  metal? 

Solution-.  The  thermal  resistivity  is  the  reciprocal  of  the 
thermal  conductivity,  and  in  the  case  of  this  transfer  resistance 
is  the  reciprocal  of  the  number  of  gram  calories  which  will  be 
transferred  to  1  square  centimeter  of  pipe  surface  from  the 
hot  gases  per  second  per  each  degree  centigrade  of  difference 
of  temperature  causing  the  flow.  This  is  a  surface  or  skin 
resistance,  and,  therefore,  no  linear  dimension  representing 
thickness  enters  into  the  calculation.  There  is  0.084  calorie 
per  second  being  transferred  to  each  square  centimeter  of  pipe, 
with  a  difference  of  temperature  acting  as  propelling  force  of 
198—10  =  188°.  The  thermal  conductivity  of  this  transfer,  k, 
is,  therefore, 

0.084 -M 88  -  0.00045 

And  the  thermal  resistivity,  R,  the  reciprocal,  viz.:  2222.  Since 
the  thermal  resistivity  of  the  iron  wall  is  l-r-0.28  =  3.57  units, 
it  follows  that  the  contact  surface  offers  2222^-3.57  =  622 
times  as  much  resistance  to  the  flow  of  heat  as  the  metal  itself. 
In  this  specific  instance,  we  can  conclude  that  the  transfer  of 
heat  from  the  gases  to  the  pipe  is  the  principal  item  which 
conditions  the  flow  of  heat,  the  passage  of  the  heat  through  the 
wall  of  the  pipe  itself  taking  place  622  times  as  readily. 

The  valuable  practical  conclusion  is  that  the  thermal  re- 
sistance of  the  walls  of  the  pipe  is  insignificant  as  compared 
with  the  whole  thermal  resistance,  and  the  thickening  or  thin- 
ning of  the  walls  of  the  pipe,  or  the  substitution  of  copper 
for  iron,  because  of  its  greater  thermal  conductivity,  is  prac- 
tically unnecessary  for  thermal  considerations,  since  such  can 
be  expected  to  make  practically  no  change  in  the  thermal  re- 
sistance of  the  whole  system. 

If  the  pipe  under  discussion,  however,  acquires  during  use  a 


206  METALLURGICAL  CALCULATIONS. 

layer  of  scale  deposited  from  the  water,  then  the  thermal  re- 
sistance of  the  pipe  or  of  the  system  is  materially  affected.  A 
study  of  the  thermal  resistivity  of  various  boiler  scales  would 
be  a  very  useful  and  practical  subject,  but  has  not  been  done,  as 
far  as  the  writer  is  aware.  Assuming  a  deposit,  0.5  centimeter 
thick,  of  material  having  the  thermal  conductivity  of  plaster  of 
paris,  for  which  k  =  0.0013,  the  specific  resistance  of  this  ma- 
terial is  0.14-J-0.0013  =  108  times  that  of  iron,  and  therefore 
0.5  centimeter  of  this  represents  54  centimeters  thickness  of 
iron.  The  thermal  resistances  in  this  case,  neglecting  that  of 
transfer  from  the  water  to  the  pipe  or  scale,  are  as  follows: 
Resistance  of  transfer,  gases  to  pipe  =  2222.0  =  85  per  cent. 

Resistance  of  0.5  c.m.  iron         =  .  '   .  =        3.6  =0         " 

0»14 

Resistance  of  0.5  c.m.  scale   =     °  \f     =    384.4  =  15 

U . UUlo 

Total  =  2610.0 

It  thus  appears  that  a  deposit  of  scale  from  the  water  to  a 
thickness  of  5  millimeters  increases  the  total  thermal  resistance 
of  the  system  some  18  per  cent,  of  its  original  amount,  and 
would  cut  down  the  efficiency  of  this  part  of  the  heating  surface 
of  a  boiler  by  this  amount.  These  considerations  are  not  only 
of  vital  importance  to  the  steam  boiler  engineer,  but  they 
are  the  essential  principles  which  condition  the  efficiency  of 
steam-heating  apparatus,  feed-water  heaters,  hot-air  stoves  and 
ovens,  air-cooling  of  parts  of  furnaces  and  the  efficiency  of 
water  jackets. 

PRINCIPLES  OF  HEAT  TRANSFER. 

We  have  already  had  to  speak  of  the  transfer  of  heat  from 
fluids  to  solids,  or  vice  versa,  and  in  one  specific  case  we  de- 
duced the  value  2222  for  the  transfer  resistivity  from  hot  gases 
to  the  surface  of  iron  pipe,  meaning  thereby  that  for  each 
degree  of  temperature  difference  between  the  gases  and  out- 
side of  the  pipe  0.00045  gram  calorie  passed  per  second  through 
each  square  centimeter  of  contact  surface.  A  consideration 
of  the  transfer  of  heat  through  such  contact  surfaces,  from 
gases  or  liquids  to  solids  and  vice  versa,  has  shown  that  the 


CONDUCTION  AND  RADIATION  OF  HEAT.  207 

transfer  resistivity  varies  with  the  solid  and  with  the  fluid  con- 
cerned, but  much  more  with  the  latter  than  with  the  former, 
and  is  very  largely  dependent  upon  the  circulation  of  the 
fluid,  that  is,  upon  the  rate  at  which  it  is  renewed,  and  there- 
fore upon  its  velocity.  The  conductivity  or  resistivity  of  such 
a  transfer  must,  therefore,  contain  a  term  which  includes  the 
velocity  of  the  fluid.  Various  tests  by  physicists  have  shown 
the  specific  conductance  (or  conductivity  of  transfer)  to  vary 
approximately  as  the  square  root  of  the  velocity  of  the  fluid. 

From  metal  to  air  or  similar  gases,  the  mean  velocity  of 
flow  being  expressed  in  centimeters  per  second,  and  the  other 
units  being  square  centimeters  and  gram  calories,  the  transfer 
resistivity  is  approximately 

36,000 

K   =   -  r=. 

2  +  Vv 

and  the  transfer  conductivity  of  the  contact 
k  =  0.000028 


From  hot  water  to  metal  the  relations  are  similar,  but  the 
conductivity  is  much  better  Experiments  show  values  as 
follows  : 

k  =  0.000028  (300  +  180\/v) 

36,000 
K.  = 


300  +  lSO\/v 

Illustration:  In  the  preceding  case  of  the  iron  pipe,  calculate 
the  difference  of  temperature  of  the  water  in  the  pipe  and  the 
inner  surface  of  the  pipe,  assuming  the  water  to  be  passing 
through  at  a  velocity  of  4  centimeters  per  second. 

Using  the  above  given  formula,  the  heat  transfer  per  1° 
difference  would  be 

0.000028  (300  +  180\/4)'  =  0.0185  calories, 
and  the  difference  to  transfer  0.084  calories  per  second  will  be 


-4°6 
0.0185 

The  inner  surface  of  the  iron  pipe  will  be,  therefore,  con- 
tinuously 4°.6  higher  than  the  water,  and,  therefore,  at  14°.6; 


208  METALLURGICAL  CALCULATIONS. 

the  outer  surface  will  be  continuously  0°.3  higher,  or  prac- 
tically at  15°. 

Illustration:  A  steam  radiator,  surface  at  about  100°  C., 
caused  a  current  of  hot  air  to  rise  having  a  velocity  of  about 
10  centimeters  per  second,  which  was  insufficient  to  keep  the 
room  warm.  An  electric  fan  was  set  to  blow  air  against  the 
radiator,  which  it  did  with  a  velocity  of  about  300  c.m.  per 
second,  and  keeping  the  room  comfortably  warm.  What  were 
the  relative  quantities  of  heat  taken  from  the  radiator  in  the 
two  cases? 

The  relative  thermal  conductivities  of  transfer  were 

2  +  VT(j  :  2  +  X/300 
or  5  : 16 

Showing  over  three  times  as  much  heat  taken  away  per  unit 
of  time  in  the  second  instance. 

This  illustration  proves  the  great  efficiency  which  the  metal- 
lurgist may  attain  in  air  cooling  of  exposed  surfaces,  by  blow- 
ing the  air  against  them  instead  of  merely  allowing  it  to  be 
drawn  away  by  its  ascensive  force. 

Problem  24. 

Dry  air  of  the  volume  of  33,000  cubic  meters  per  hour  passes 
through  an  iron  pipe  exposed  to  the  air,  30  meters  long,  1.5 
meters  inside  diameter,  thickness  of  walls  1  centimeter,  and 
lined  inside  with  5  centimeters  of  fire-brick.  Assume  the  hot 
air  entering  at  1,000°,  the  outside  air  to  be  at  3°,  the  coefficient 
of  internal  transfer  0.000028  (2  +  Vv),  the  conductivity  of 
the  fire-bricks  0.0014,  of  the  iron  0.14,  of  external  transfer 
0.000028  (2  +  x/v7,  the  vel°citv  °f  tne  win<l  against  the  outside 
10  kilometers  per  hour. 

Required:    The  temperature  of  the  hot  air  leaving  the  tube. 

Solution:  The  mean  temperature  in  the  tube  is  the  factor 
which  conditions  the  mean  velocity  in  the  tube,  and  the  rate  of 
flow  of  heat  towards  the  outside.  If,  therefore,  we  let  t  repre- 
sent that  mean  temperature,  the  solution  can  be  stated  in  the 
simplest  terms.  We  then  have: 

Volume  of  air  per  hour,  at  0°,     =  33,000  cubic  m. 
Volume  of  air  per  second,  at  0°,  =    9.167 


CONDUCTION  AND  RADIATION  OF  HEAT.  209 

t  +  273 

Volume  of  air  per  second,  at  t,    =   9.167  —  _ 

Z7o 

Cross-section  of  inside  of  tube 

(1.5—  0.1)2X0.7854  =  1.54  sq.  m. 

Velocity  of  air  at  t,  in  pipe 


=  5.95  (1  +  at)  m.  p.  s. 
=  595   (1  +  at)  c.m.  p.  s. 
Coefficient  of  internal  transfer     = 
0.000028 


Coefficient  of  conductance  of  5  c.m.  of  fire-brick  lining  = 
0.0014  -T-  5  =  0.00028 

Coefficient  of  conductance  of  1  c.m.  of  iron  = 
0.14-^1.=  0.14 

Coefficient  of  external  conductance  (v  =  278  c.m.  per  sec.)  = 

0.000028  (2  +  X/278)"  =  0.000524 
Total  fall  of  temperature  of  air   =  2  (1000—  t) 

End  temperature  of  the  air          =  1000  —  2  (1000  —  t) 

=  2  t—  1000 

Mean  specific  heat  of  air  per  cubic  meter  between  1000  and  end 
temperature  = 

0.303  +  0.000027  (2t) 

Heat  given  out  by  the  air  per  hour  = 

33,000X2(1000—  t)X  (0.303  +  0.000027  (2t)  )  = 

19,998,000—16,434  1—  3.564  ta 

This  heat  is  the  quantity  transferred  through  the  pipe  in 
kilogram  Calories.  The  outside  surface  of  the  pipe  may  be 
taken  as  the  conducting  surface,  because  the  larger  part  of  the 
resistance  to  the  flow  of  heat  takes  place  there.  If  we  desired 
to  be  more  exact,  the  mean  area  of  iron,  fire-bricks  and  the 
inner  surface  could  be  each  used  separately.  The  outside 
diameter  being  150  centimeters,  and  the  length  30  meters,  the 
outer  surface  is  1.52X3.1416X30=142.3  square  meters  = 
1,423,000  square  centimeters,  the  total  driving  force  is  the 
inside  temperature  minus  that  outside,  or  t  —  3,  and  the  total 


210  METALLURGICAL  CALCULATIONS. 

thermal  resistance  is  the  sum  of  the  four  thermal  resistances  in 
series,  that  is,  the  sum  of  thermal  resistance  gas  to  fire-bricks  =» 


(14- a  t) 
thermal  resistance  of  fire-brick  lining  = 

1 
0.00028 

thermal  resistance  of  iron  shell  = 

1 

0.14 

thermal  resistance  of  iron  to  air  = 

1 
0.000524 

The  reciprocal  of  the  sum  of  these  four  resistances  is  the 
thermal  conductance  of  the  system  per  square  centimeter, 
which,  multiplied  by  the  conducting  surface,  1,423,000  square 
c.m.,  and  by  the  total  difference  of  temperature,  t — 3,  gives 
the  heat  transferred  per  second  in  gram  calories. 

We,  therefore,  have  the  final  equality  expressed  as  the  heat 
given  up  by  the  air,  per  second,  equals  the  heat  transmitted  to 
the  outside  air  per  second;  i.e., 

(19,998,000— 16.434 1— 3.564  tz)  =- 


0.000028  (2  + V595  (l  +  «t)        0.00028       0.14         0.000524 

XI, 423,000 X  (t— 3) 
Whence  t  =  965°.5 

And  the  temperature  of  the  air  at  the  end  of  the  tube  is 

2t— 1000  =  931°  (1) 

It  would  be  interesting  to  compare  the  temperatures  of  the 
inside  and  outer  surfaces  of  the  tube.  The  air  inside  is  at  a 
mean  temperature  of  965°,  the  heat  transmitted  per  second  is 
0.1574-gram  calories  per  square  centimeter  of  outer  surface, 


CONDUCTION  AND  RADIATION  OF  HEAT.  211 

whose  thermal  conductivity  is  0.000515,    and  therefore,  the  out- 
side surface  must  be 

0.1574 


0.000515 

hotter  than  the  surrounding  air;  that  is,  must  be  at  309°  C. 

The  extra  loss  of  heat  by  radiation  from  this  outside  surface 
would  be  small  at  that  low  temperature,  but  has  not  been 
allowed  for  in  the  working  of  the  problem. 

For  such  metallurgical  problems  we  need  the  data  as  to  the 
thermal  conductivity  or  resistivity  of  ordinary  furnace  ma- 
terials. These  have  been  determined  in  but  few  instances,  and 
in  most  cases  not  at  the  high  temperatures  at  which  they  are 
practically  used.  A  whole  series  of  physico-metallurgical  ex- 
periments is  needed  just  upon  this  point.  The  following  are 
probably  nearly  all  that  have  been  determined  and  the  values 
published.  The  unit  k  is  the  C.  G.  S.-gram  calories  unit,  the 
same  as  used  for  the  metals. 

k. 
Ice  (datum  useful  in  refrigerating  plants,  where  pipes 

become  coated  with  ice,  as  in  Gayley's  method  of 

drying  blast) 0.00500 

Snow 0.00050 

Glass  (10°— 15°) 0.00150 

Water 0.00120 

Quartz  sand  (18°— 98°) 0.00060 

Carborundum  sand  (18°— 98°) 0.00050 

Silicate  enamel  (20°— 98°) . .' 0.00040 

(Explains  the  small  conductance  of  enameled  iron 

ware.) 

Fire-brick,  dust  (20°— 98°) 0.00028 

Retort  graphite  dust  (20°— 100°) 0 . 00040 

(Datum   useful   where   articles   are   packed   in   this 

poorly  conducting  material.) 

Lime  (20°— 98°) 0.00029 

(Datum  would  be  highly  useful  for  oxyhydrogen 
platinum  furnaces,  if  it  were  only  known  at  high  tem- 
peratures.) 

Magnesia  brick,  dust  (20°— 100°) 0 . 00050 

Magnesia  calcined,  Grecian,  granular  (20° — 100°) 0.00045 

Masonry  0.0036  to  0.0058 

Water,  uncirculated         0.0012  to  0.0016 


212  METALLURGICAL  CALCULATIONS. 

Magnesia  calcined,  Styrian,  granular  (20°— 100°) 0.00034 

Magnesia  calcined,  light,  porous  (20°— 100°) 0.00016 

Infusorial  earth  (Kieselguhr)  (17°— 98°) 0.00013 

Infusorial  earth  (0°— 650°) 0.00038 

Clinker,  in  small  grains  (0°— 700°) 0.00110 

Coarse  ordinary  brick  dust  (0°— 100°) 0.00039 

Chalk  (0°— 100°) 0.00028 

Wood  ashes  (0°— 100°).  .  .  : 0.00017 

Powdered  charcoal  (0°— 100°) 0.00022 

Powdered  coke  (0°— 100°) 0.00044 

Gas  retort  carbon,  solid  (0°— 100°) .0.01477 

Cement  (0°— 700°) f 0.00017 

Alumina  bricks  (0°— 700°) 0.00204 

Magnesia  bricks  (0°— 1300°) 0.00620 

Fire-bricks  (0°— 1300°) 0.00310 

Fire-bricks  (0°— 500°) 0.00140 

Marble,  white  (0°) 0.0017 

Pumice 0.0006 

Plaster  of  paris 0.0013 

Felt 0.000087 

Paper .0.00040 

Cotton , 0.000040 

Wool 0.000035 

Slate 0.00081 

Lava 0.00008 

Pumice 0.00060 

Cork 0.00072 

Pine  wood ' 0.00047 

Oak  wood 0.00060 

Rubber 0.00047 

A  study  of  the  above  table  will  in  many  cases  show  the 
metallurgist  how  conduction  of  heat  can  be  checked,  and  to 
what  degree.  The  substance  chosen  must  be  able  to  stand  the 
temperature  without  being  destroyed;  but  it  is  in  many  cases 
possible  to  use  one  kind  of  material  inside,  where  the  heat  is 
greatest,  and  a  material  of  much  poorer  conductivity  outside, 
where  the  heat  will  not  destroy  it.  Such  compound  linings  or 
coverings  may  be  very  advantageous.  Infusorial  earth  is  one 
of  the  very  best  insulators  for  moderate  temperature;  above 


CONDUCTION  AND  RADIATION  OF  HEAT. 


213 


a  bright  red  it  loses  efficiency  greatly,  and  is  then  hardly  better 
than  powdered  fire-brick. 

Mr.  Irving  Langmuir  has  recently  conducted  important  ex- 
periments on  convection  and  radiation  of  heat,  recorded  in  a 
paper  before  the  American  Electrochemical  Society  (Trans- 
actions (1913)  23,  299-332).  His  experimental  results  for 
convection  alone,  from  horizontal  and  vertical  surfaces,  to  still 
air  at  27°  C.,  are  as  follows,  first  in  calories  per  second  and  next 
in  watts: 

GRAM-CALORIES  PER  SEC.  PER  SQ.  CM. — TO  STILL  AIR  AT  27°  C. 


100° 

200° 

300° 

400° 

500° 

Downward  from  - 
from 
Upward        from  - 

—  Surface 
Surface 
-  Surface 

0.002 
0.005 
0.011 

0.014 
0.029 
0.032 

0.024 
0.050 
0.055 

0.035 
0.074 
0.081 

0.047 
0.099 
0.108 

WATTS  PER  SQ.  CM. — TO  STILL  AIR  AT  27°  C. 


100° 

200° 

300° 

400° 

500° 

Downward  from  - 
from 
Upward        from  — 

—  Surface 
Surface 
-  Surface 

0.010 
0.020 
0.046 

0.060 
0.120 
0.132 

0.100 
0.210 
0.233 

0.146 
0.308 
0.341 

0.196 
0.416 

0.452 

These  figures  are  for  convection  alone,  and  do  not  include 
radiation.  They  are  different  from  the  formula  given  on  page 
207,  but  are  probably  more  reliable.  If  data  are  required  for 
over  500°,  the  above  values  may  be  plotted  and  the  curves 
extrapolated. 

If  the  air  is  in  motion,  Mr.  Langmuir  found  that  the  heat  loss 

by  convection  could  be  expressed  as  -\/  times  the  loss  to 

\      oo 

still  air,  where  V  is  the  velocity  of  air  current  in  centimeters  per 
second.  Therefore,  the  velocity  being  known,  evaluate  the 
above  expression  and  multiply  the  tabulated  values  by  this 
factor. 

RADIATION. 

A  body  placed  in  a  vacuum,  with  no  ponderable  substance  in 
contact  with  it,  radiates  heat  to  its  surroundings.  All  experi- 
ments so  far  made  confirm  the  accuracy  of  Stefan's  law,  that 
every  hot  body  radiates  energy  in  proportion  to  the  fourth 
power  of  its  absolute  temperature,  and  that  the  transfer  of 
heat  by  radiation  is  therefore  proportional  to  the  difference 


214  METALLURGICAL  CALCULATIONS. 

between  the  fourth  powers  of  the  absolute  temperatures  of  the 
hot  body  and  its  surroundings,  respectively. 

Mr.  Langmuir  uses  Stefan's  law  of  radiation  expressed  in 
the  following  form 

QR  =  E  5.9   X  10-12(T24— TV)  watts  per  sq.  cm. 

=  E  1.41X10-12(T24— TV)  cal.  per  sec.  per  sq.  cm. 

These  formulae  express  the  fact  that  the  radiation  from  a  body 
at  absolute  temperature  T2  to  one  at  7\,  is  proportional  to  the 
difference  between  the  fourth  powers  of  the  temperatures  in 
question,  and  that  for  each  unit  of  such  difference  the  radiation 
loss  from  the  hotter  body  is  1.41X1Q-12  calories  per  square 
centimeter  per  second,  if  the  body  has  unfi;y  radiating  power 
(black  body),  or  Exl.41  XlO~12  calories  if  its  coefficient  ot 
radiating  power  (emissivity)  is  E.  If  t1.c  heat  flow  is  expressed 
as  energy  flow,  1 . 41  calories  per  second  =5.9  watts. 

For  convenience  sake,  to  avoid  large  numbers,  the  absolute 
temperatures  may  be  expressed  on  a  scale  of  1000°,  in  which 
case  10~12  disappears,  and  the  formula  becomes: 


=  EX1.41 


viooo/    watts 


1000/       UOOO 


cal.  per  sec. 


In  the  above  form,  this  formula  is  very  convenient  for  cal- 
culating radiation  losses  from  any  body  at  absolute  temperature 
T2  to  its  surroundings  at  rb  provided  the  emissivity  of  the  hot 
body,  E,  is  known.  Dull  lampblack  has  an  emissivity  nearly 
1 . 00,  but  all  other  materials  have  emissivities  less  than  unity, 
the  polished  metals  having  values  as  low  as  0.02.  Unfortu- 
nately, total  energy  emissivity  is  not  quite  independent  of 
temperature,  and  most  of  the  determinations  have  been  made 
for  low  temperatures  only. 

In  the  following  table,  some  emissivities  have  been  calculated 
from  the  older  experiments  on  radiation,  and  some  are  Mr. 
Langmuir 's  own  values. 


CONDUCTION  AND  RADIATION  OF  HEAT. 


215 


Radiating  Capacity  (Emissivity) 


Theoretical  black  body. 
Lampblack  

(At  all  tempe 
(50°)  0.95  - 
(50°)  0.68 
(50°)  0.64 
(50°)  0.56 
(50°)  0.52 
(50°)  0.17 
(1200°)  0.28 
(1600°)  0.28 
(500°)  0.85 
(440°)  0.91 
(50°)  0.56 
(50°)  0.47 
(5001  0.11 
(50°)  0.04 
(500°)  0.98 
(50°)  0.99 
(50°)  0.60 
(50°)  0.65 
(50°)  0.90 
(50°)  0.77 
(50°)  0.39 
(50°)  0.03 
(1075°)  0.16 
(240°)  0.156 
(50°)  0.67 
(50°)  0.04 
(50°)  0.02 
(50°)  0.03 
(50°)  0.33 
(50°)  0.035 
(1160°)  0.14 
(50°)  0.07 
(50°)  0.50 
(50°)  0.50 
(50°)  0.11 
(50°)  0.04 
(50°)  0.05 
(50°)  0.35 
(800°)  0.68 
(50°)  0.04 
(50°)  0.07 
(50°)  0.72 
(3250°)  0.39 
(500°)  0.10 
(50°)  0.06 
(50°)  0.06 
(50°)  0.10 
(50°)  0.49 
(50°)  0.25 
(300°)  0.73 
(50°)  0.60 
(50°)  0.60 
(50°)  0.60 

ratures)  =  l.( 
-  0.98 

(300)°  0.67 

(1750°)  0.28 

(1000°)  0.88 
(468°)  0.85 

(1000°)  0.99 
(720°)  0.60 

(300°)  0.82 
(300°)  0.70 
(300°)  0.26 
(1000°)  0.09 
(1175°)  0.15 
(360°)  0.19 
(127°)  0.45 

(300°)  0.03 
(500°)  0.07 
(126°)  0.37 
(300°)  0.061 
(1430°)  0.16 

(300°)  0.30 
(300°)  0.41 
(500°)  0.19 

(500?)  0.06 
(300°)  0.40 
(1000°)  0.75 

(300°)  0.14 
(1000°)  0.72 
(Langmuir)  . 

(500°)  0.09 
(500°)  0.09 
(500°)  0.40 

(500°)  0.50 
(400°)  0.79 

)0 

(500)°  0.53 

(Thwing) 
(1200°)  0.89 

(945C)  0.60 

(400^)  0.92 
(500°)  0.73 
(500°)  0.23 

(1275°)  0.13 

(500°)  0.04 

(500°)  0.08 
(1695°)  0.175 

(500°)  0.38 
(1000°)  0.37 

(1000°)  0.17 
(500°)  0.48 
(1075)°  0.80 

(500°)  0.23 

(1170°)  0.61  (Burgess) 
(500°)  0.95 

(1000°)  0.70 
(1000°)  0.20 

(1290°)  0.1  5  Thwing 

(1000°)  0.124 
(Burgess) 

(710°)  0.62 
(1300°)  0.88 

Smoothblack  paint  
Black  paper  
Cast  iron,  rusted  
Cast  iron,  new  

Cast  iron,  polished  

Soft  steel,  liquid  
Iron  oxide  scale  
Iron  rust  

Russian  sheet  iron  
Ordinary  sheet  iron  
Leaded  sheet  iron  
Tinned  sheet  iron  
Hematite  

Chromite  

Cuprous  oxide  

Graphite,  mat  

Copper,  oxidized  
Copper,  calorized  
Copper,  polished  
Copper,  liquid        

Aluminium  polished.  .  .  . 
Aluminium  paint  

Brass,  polished  

Gold,  polished  
Gold,  enamel  
Platinum,  polished 

Silvered  paper  

Monel  metal,  bright.  .  .  . 
Monel  metal,  oxidized.  . 

Zinc,  bright.. 

Nickel,  bright  

Nickel  oxide  

Tin   bright 

Magnesium. 

Silicon  

Alumina  

Magnesia  

Lime.. 

Glass 

Porcelain. 

White  fire  clay. 

Building  stone   . 

Plaster  

Wood. 

APPENDIX  TO  PART  I. 

Problem  25. 

(1)  Write  the  equations  showing  the  relative  weights  and 
relative  volumes  (of  gases)  concerned  in  the  combustion  of 

Methane  (marsh  gas) C   H4 

Acetylene C2  H2 

Ethylene  (olefiant  gas) C2  H4 

Methylene C2  H6 

Allylene C3  H4 

Propylene C3  H8 

Propylene  hydride C3  H8 

Benzine C6  H8 

Turpentine  (liquid) C10H18 

Napthaline  (liquid) C10H8 

(2)  The  molecular  heats  of  combustion  of  the  above  gases  to 
CO2  and  liquid  H2O  are  given  by  Berthelot  as: 

for  C   H4 213,500  Calories 

"  C2  H2 315,700 

"  C2  H4 341,100 

«  C2He 372,300        * 

"  C3  H4 473,600        « 

"  C3  H6 499,300        " 

"  C3  H8 528,400 

"  C8H6 784,100 

«  CioHie  (liquid) 1,490,800 

«  CioHs  (liquid) 1,241,800        * 

(a)  Calculate  the  molecular  heats  of  combustion  to  CO2  and 
H2O  vapor. 

(b)  Calculate  the  molecular  heats  of  formation  of  the  hydro- 
carbons themselves,  using  (C,  O2)  97,200  and  (IPO)  69,000  liquid 
or  58,060  gas. 

(c)  Calculate  the  heat  of  combustion,  to  CO2  and  H20  vapor,  of 
one  cubic  meter  of  each  gas,  and  of  one  cubic  foot  in  B.  T.  units. 

217 


218 


METALLURGICAL  CALCULATIONS. 


Answers  (1): 

i 

ii 

i 

ii 

CH4 

+     202 

=     CO2       + 

2H2O 

16 

64 

44 

36 

ii 

V 

IV 

ii 

2C2H2 

+     502 

=     4CO2      + 

2H2O 

52 

160 

176 

36 

i 

in 

ii 

ii 

C2H4 

+     3O2 

=      2CO2     + 

2H2O 

28 

96 

88 

36 

ii 

VII 

IV 

VI 

2C2H6 

+      7O2 

=      4CO2     + 

6H2O 

60 

224 

176 

108 

i 

IV 

in 

ii 

C3H4 

+     402 

PP      3C02    + 

2H20 

40 

128 

132 

36 

ii 

IX 

VI 

VI 

2C3H6 

+      9O2 

=      6CO2    + 

6H2O 

84 

288 

264 

108 

i 

V 

in 

IV 

C3H* 

+     5O2 

=      3C02     + 

4H2O 

44 

160 

132 

72 

i 

XIV 

X 

VIII 

C10H16 

+     14O2 

=      10CO2   + 

8H2O 

136 

448 

440 

144 

i 

XII 

X 

IV 

C10H8 

+      12O2 

=      10C02   + 

4H2O 

128 

384 

440 

72 

(2) 

Molecular  Heat 

Molecular  Heat 

Heat  of  Combustion  of 

of  Formation 

of  Combustion 

1  m3  (Cal.) 

if*.3 

(C 

Amorphous.) 

(to  H*O  Vapor.) 

B.T.U. 

C  H4 

+  21,700 

+     191,620 

+   8,623 

+    970 

C2H2 

—52,300 

+    304,760 

+  13,714 

+  1,543 

C2H4 

—  8,700 

+    319,220 

+  14,365 

+  1,616 

C2H6 

+  29,100 

+    339,480 

+  15,277 

+  1,719 

C3H4 

—44,000 

+    451,720 

+  20,327 

+  2,287 

C3H6 

—      700 

+    466,480 

+  20,992 

+  2,362 

C3H8 

+  39,200 

+    486,640 

+  21,899 

+  2,464 

C6H6 

+   6,100 

+    751,280 

+  33,808 

+  3,803 

CioHie 

(liquid) 

+  33,200 

+  1,403,280 

.... 

.... 

C10H16 

(gas) 

+  23,800 

+  1,412,680 

+  63,570 

+  7,152 

C10H8 

(liquid)     • 

+   6,200 

+  1,198,040 

.... 

.... 

APPENDIX.  219 

Problem  26. 

The   following  typical   analyses   are   from    Poole's   book  on 
11  The  Calorific  Power  of   Fuels  ": 

Bituminous  Coal.          Fuel  Oil .  Natural  Gas.^ 

"Carnegie,  Pa."        "Lima,  O."  "Findlay,  0." 

Carbon  77.20  80.2                     H2  1.64 

Hydrogen  5.10  17.1                     CH4  93.35 

Oxygen  7.22  1.3                     C2H4  0.35 

Nitrogen  1.68  1.4                     CO2  0.25 

Sulphur  1.42  CO  0.41 

Moisture  1.45  O2  0.39 

Ash  5.93  N2  3.41 

H2S  0.20 

Required:    (1)  The  practical,  metallurgical  calorific  powers  of 

(a)  The  coal,  per  kilogram  or  pound  7,447 

(b)  The  oil,  per  kilogram  or  pound  11,393 

(c)  The  gas,  per  cubic  meter  8,143  kg.  Cal. 

(d)  The  gas,  per  1000  cubic  feet  509,000  Ib.  Cal. 
(2)  Compare  the  calorific  values  of  1  ton  (2240  Ibs.)  of  coal, 

1  ton  (2000  Ibs.)  of  oil,  and  1000  cubic  feet  of  natural  gas. 

32.8:44.8:1. 
Problem  27. 

The  composition  of  some  commercial  gases  used  as  fuels  is 
given  by  Wyer  (Treatise  on  Producer-Gas  and  Gas  Producers, 
p.  50),  as  in  following  table: 

H2     CH4     C2H4       N2      CO       O2      CO2 

Natural    gas    (Pitts- 
burg) 3.0     92.0       3.0       2.0 

Oil  gas 32.0     48.0     16.5       3.0  0.5 

Coal    gas,    from    re- 
torts  46.0     40.0       5.0       2.0       6.0     0.5     0.5 

Coke-oven  gas 50.0     36.0       4.0       2.0       6.0     0.5     1.5 

Carburetted  water- 
gas. . 40.0     25.0       8.5       4.0     19.0     0.5     3.0 

Water  gas 48.0       2.0  5.5     38.0     0.5     6.0 

Producer   gas    (hard 

coal,  using  steam)  20.0  49.5     25.0     0.5     5.0 

Producer    gas    (soft 

coal) 10.0       3.0      0.5     58.0     23.0    0.5     5.0 


220  METALLURGICAL  CALCULATIONS. 

Requirements:    For  each  of  the  above  gases  calculate 

(1)  The  volume  of  air  theoretically  necessary  to  burn  it. 

(2)  The  volume  of  the  products  of  combustion. 

(3)  The  calorific  power 

(a)  per  cubic  meter,  in  kilogram  Calories. 

(b)  per  cubic  foot,  in  pound  Calories. 

(c)  per  cubic  foot,  in  British  Thermal  Units. 

(4)  The  theoretical  temperature,  if  burned  cold  with  cold 
air  (calorific  intensity). 

7?/>c*,/7/c'                                        Volume  Volume 

J\VWU*.  of  air  to  of  prod-  Calorific  Power.  Calori- 

burnl  ucts  per  Kg.  Cal.  Lb.  Cal.  B.T.U.  fie  In- 

volume.  1  volume  per  m*.  per  ft*.  per  ffi.  tensity. 

Natural  gas  (Pittsburg)  9 . 35  10.33  8423  526  948  1852° 

Oil  gas : 7.74  8.58  7350  460  828  1915° 

Coal  gas,  from  retorts . .  5 . 79  6 . 50  5550  347  625  1896° 

Coke-oven  gas 5 . 36  6 . 08  5159  322  579  1892° 

Carburetted  water-gas..  5. 06  5.77  5008  313  563  1914° 

Water  gas 2.24  2.81  2590  162  292  1928° 

Producer  gas  (with 

steam) 1.08  1.86  1290  81  146  1676° 

Producer  gas  (ordinary)  1.13  1.98  1300  82  148  1555° 

Problem  28. 

An  anthracite  coal  containing  on  analysis 

Carbon 89  per  cent. 

Hydrogen 3 

Oxygen 1 

Ash 7 

is  burned  under  a  boiler,  and  the  ashes  produced  weigh,  dry, 
10  per  cent,  of  the  weight  of  the  coal  used.  The  chimney  gases, 
dried  and  then  analyzed,  contain  CO2  15.3  per  cent.,  O2  3.5 
per  cent.,  N2  81.2  per  cent.,  and  25.9  grams  of  water  is  obtained 
for  each  cubic  meter  of  dry  gas  collected. 

Required:     (1)  The  volume  of  chimney  gas  (measured  dry) 
produced  per  kilo,  of  coal  burned  10.41  m3. 

(2)  The  volume  of  air  used,  measured  dry  at  normal  condi- 
tions, per  kilo,  of  coal  burned  10.67  ms. 

(3)  The  weight  of  dry  air  used  13.80  kg. 

(4)  The  volume  of  dry  air  theoretically  necessary  for  the 
complete  combustion  of  one  kilo,  of  coal  8.72  m3. 


APPENDIX.  221 

(5)  The  excess  of  air  used,  in  percentage  of  that  theoretically 
necessary  for  the  perfect  combustion  of  the  coal  22.36%. 

(6)  The  volume  of  chimney  gas,  at  standard  conditions,  per 
kilo,  of  coal  burned  10.74  m3. 

Problem  29. 

An  anthracite  coal  contains 

Carbon 89  per  cent. 

Hydrogen 3         " 

Oxygen 1 

Ash 7 

It  is  burned  on  a  grate,  using  15  per  cent,  more  air  than  is 
theoretically  needed  for  its  perfect  combustion.  The  ashes 
weigh  10  kilos,  per  100  kilos,  of  coal  burned. 

Required:  (1)  The  volume  of  air  (assumed  dry  and  at  standard 
conditions)  used  per  kilo,  of  coal  burned  9.71  m3. 

(2)  The  number  of  cubic  feet  of  air  per  pound  of  coal      155.4. 

(3)  The  percentage  composition   (by  volume,  of  course)  of 
the  chimney  gas,  assuming  it  contains  no   soot   or  unburned 
gas.  N2  77.9,  CO2  16.1,  O2  2.6,  H2O  3.4. 

(4)  The  percentage  composition  of  the  same,  if  first  dried 
and  then  analyzed  N2  80.6,  CO2  16.7,  O2  2.7. 

(5)  The  number  of  grams  of  moisture  carried  per  cubic  meter 
of  dried  gas  measured  28.4. 

(6)  The  number  of  grains  per  cubic  foot  12.4. 

(7)  The  volume  of  the  chimney  gases,  at  standard  conditions, 
(water  assumed  uncondensed)  per  kilo,  of  coal  burned     9.85  m3. 

(8)  The  number  of  cubic  feet  of  products  per  pound  of  coal 

157.6. 

(9)  The  volume  of  the  products  in  cubic  meters  at  350°  C. 
and  700  m.m.  pressure  24.4. 

(10)  The  volume  of  the  products  in  cubic  feet  at  600°  F.  and 
29  inches  pressure  349.6. 

Problem  30. 

A  bituminous  coal  contains 

Carbon 75  per  cent. 

Hydrogen 5         " 

Oxygen 10 

Ash..  .10 


222  METALLURGICAL  CALCULATIONS. 

It  is  powdered  and  injected  into  a  cement  kiln  with  25  pet 
cent,  more  air  than  is  theoretically  necessary  for  its  complete 
combustion.  The  kiln  calcines  200  barrels  of  cement  daily, 
using  125  pounds  of  coal  per  barrel  of  cement.  Assume  outside 
air  at  0°  C.  Hot  gases  enter  stack  at  819°  C.  Omit  from  cal- 
culation the  water  vapor  and  carbonic  acid  gas  expelled  from 
the  charge. 

Required:  (1)  The  volume  of  air  per  minute  which  the 
blower  or  fan  must  furnish.  2,672  ft3. 

(2)  The  volume  of  the  hot  products  of  combustion  per  min- 
ute, as  they  pass  into  the  stack.  11,044  ft3. 

Problem  31. 

A  bituminous  coal  (as  of  Problem  30)  contains 

Carbon 75  per  cent. 

Hydrogen 5 

Oxygen..., 10 

Ash 10 

It  is  burned  under  a  boiler  with  25  per  cent,  more  air  than 
is  theoretically  necessary  for  its  complete  combustion.  As- 
sume combustion  perfect,  and  no  unburnt  carbon  to  remain 
in  the  ashes'.  The  wet  steam  produced  contains  3  per  cent, 
of  water  of  priming,  and  8.82  pounds  of  water  is  evaporated  per 
pound  of  "coal  burnt.  Steam  pressure  6  atmospheres  effective 
pressure.  Outside  air  0°  C.,  feed  -water  10°  C.,  stack  gases 
310°  C. 

Required:  (1)  The  calorific  power  of  the  coal,  by  calculation, 
water  formed  considered  uncondensed,  in  pound  calories  per 
pound  of  coal,  and  British  Thermal  Units  per  pound.  7096;  12773. 

(2)  The  percentage  of  the  calorific  power  of  the  coal  repre- 
sented by  the  heat  in  the  dry  steam  produced  78.0. 

(3)  ditto  in  the  water  of  priming  0.6. 

(4)  ditto  in  the  sensible  heat  of  the  chimney  gases          14.5. 

(5)  ditto  lost  by  radiation  and  conduction  6.9. 

(6)  What    increase    in    the    net    efficiency    (requirement    2) 
would  be  obtained  by  using  a  feed  water  heater,  which  ab- 
stracted 60  per  cent,  of  the  sensible  heat  in  the  chimney  gases, 
all  other  conditions  remaining  constant?  8.6%. 


APPENDIX.  223 

Problem  32. 
A  bituminous  coal  contains  (as  in  Problems  30  and  31): 

Carbon 75  per  cent. 

Hydrogen , 5 

Oxygen ' 10 

Ash 10 

It  is  blown  into  a  revolving  cylindrical  furnace  with  a  high 
pressure  blast  which  supplies  only  14.9  percent,  of  the  air  neces- 
sary for  its  complete  combustion,  producing  near  the  burner  a 
highly  luminous  flame  surrounded  by  a  cylindrical  sheath  of 
auxiliary  air  drawn  in  by  the  injector  action  of  the  fuel-air 
jet.  Assuming  that  in  the  body  of  the  jet  the  hydrogen  only 
of  the  coal  is  consumed,  the  luminosity  being  due  to  un con- 
sumed particles  of  carbon,  and  ash;  that  the  mean  specific  heat 

120 

of  carbon  is  0.5 ,  of  the  ash  0.25 

t 

Required:  (1)  The  theoretical  temperature  of  the  interior 
of  the  jet  of  burning  fuel  1140°C. 

(2)  The  temperature  if  the  supply  of  air  in  the  fuel  jet  is 
increased  to  that  theoretically  necessary  for  perfect  combustion 
of  the  whole  fuel  .  1945°  C. 

Problem  33. 

Calculate  the  maximum  temperature  obtainable  in  the  region 
of  the  tuyers  of  a  blast-furnace. 

(1)  Using  cold,  dry  air, 1683°  C. 

(2)  Using  dry  air,  heated  to  700°  C.  2272°  C. 

(3)  Using  moist  air,  heated  to  700°  C.,  the  amount  of  moisture 
present  being  such  as  air  at  37°  C.  can  carry  when  saturated 
with  moisture.  1947°  C. 

(4)  Using  cold,  moist  air  of  above  composition.  1367°  C. 

Problem  34. 

Powdered  coal  having  the  following  composition  is  burnt  in  a 
cement  kiln: 

Carbon 73.60  per  cent. 

Hydrogen 5.30 

Nitrogen 1.70 

Sulphur. . . . 0.75 


224  METALLURGICAL  CALCULATIONS. 

Oxygen 10.00  per  cent. 

Ash 8.05 

Moisture 0.60 

The  finely-ground  coal  is  burnt  by  cold  air,  at  20°  C.,  and 
760  m.m.  pressure.  The  gases  resulting,  together  with  carbon 
dioxide  gas  from  the  charge,  pass  into  the  stack  at  820°  C. 
The  analysis  of  the  flue  gases,  by  volume,  dried  before  analysis, 
is: 

Carbon  dioxide 25.9  per  cent. 

Oxygen 3.1 

Carbon  monoxide 0.2  .      " 

Sulphur  dioxide not  determined. 

Nitrogen difference. 

Of  the  carbon  dioxide  in  the  gases  assume  40  per  cent,  of  it 
to  come  from  the  carbonates  in  the  charge  being  treated,  and 
not  from  the  coal.  Assume  no  water  in  the  furnace  charge. 
The  air  used  is  saturated  with  moisture,  at  20°  C.,  tension  of 
the  moisture  22  m.m.  of  mercury. 

Required:     (1)  The  theoretical  calorific  power  of  the  coal    7090. 

(2)  The  theoretical  temperature  of  the  hottest  part  of  the 
flame  1593°  C. 

(3)  The  proportion  of  the  calorific  power  of  the  fuel  carried 
out  by  the  hot  gases  50.4%. 

(4)  The  percentage  excess  of  air  admitted  above  that  theoret- 
ically required  16.4%. 

Problem  35. 

Calculate  the  draft  of  a  chimney,  in  inches  of  water  gauge 
and  feet  of  cold  air,  if  measured  at  its  base,  its  height 
being  120  feet,  inside  diameter,  (round)  6  feet,  temperature  of 
gases  inside  at  bottom  300°  C.,  at  top  200°  C.,  specific  gravity 
(air  =  1)  1.03,  and  taking  in  200  cubic  feet  (measured  at  0° 
and  760  m.m.)  of  products  of  combustion  per  second.  Section 
inside  uniform  top  to  bottom,  sides  fairly  smooth,  assume 
K  =  0.04.  Outside  temperature  0°  C.  0.91  in.;  59.25  ft. 

Problem  36. 

A  copper  cylinder  weighing  22.092  grams  was  placed  in  a 
small  iron  box  on  the  end  of  a  rod,  and  held  several  minutes 
in  the  hot  blast  main  of  a  blast-furnace.  On  removal,  and  drop- 


APPENDIX.  225 

ping  instantly  into  a  calorimeter,  containing  301.3  grams  of 
water,  the  temperature  rose  4°.  183  C.  in  three  minutes.  From 
previous  experiments  with  this  calorimeter,  it  was  known  that 
in  three  minutes  it  would  abstract  from  the  water  30  gram 
calories  for  each  1°  observed  rise  of  temperature  above  starting. 
The  mean  specific  heat  of  copper  between  0°  and  t°  being 
0.09393 +  0.00001778t,  what  was  the  temperature  of  the  hot- 
blast? 

[N.B. — Calculate  what  would  theoretically  have  been  the 
temperature  of  the  water  of  the  calorimeter  if  no  heat  had  been 
lost  to  the  calorimeter,  and  work  out  using  this  as  the  final 
temperature  of  the  water.]  618°  C. 

Problem  37. 

A  piece  of  tin-stone  (Cassiterite,  SnO2)  from  Bolivia  was 
tested  to  obtain  its  specific  heat.  It  was  heated  to  two  tem- 
peratures determined  by  a  Le  Chatelier  thermo-electric  pyro- 
meter, and  dropped  into  a  calorimeter.  Corrections  for  calori- 
meter losses  were  made  as  explained  in  the  Journal  of  the  Frank- 
lin Institute,  August,  1901.  Data  of  the  two  tests  were  as 
follows : 

Experiment  1.     Experiment  2. 

Weight  of  tin-stone  used 12.765  gms.        12.765  gms. 

Temperature  of  same 476°  C.  1018°  C. 

Weight  of  water  in  calorimeter. .  .299.4  gms.  300.7  gms. 

Temperature  of  same,  starting 16°.527  18°.705 

Temperature  of  same,  3  minutes. .  18°. 444  22°. 985 

Water  value  of  calorimeter  per  3', 

for  each  1°  rise 30  cal.  30  cal. 

Required:  A  formula  of  the  form  Sm  =  a  +  /h,  for  the  particu- 
lar piece  of  Cassiterite  used.  0.1050 +0.000006 It 

Problem  38. 

The  chimney  gases  from  a  boiler  enter  the  stack  at  a  tem- 
perature of  400°  C.  Their  composition  is: 

Carbon  dioxide 15.0  per  cent. 

Oxygen 5.9 

Nitrogen 79.1 

What  percentage  of  the  total  calorific  power  of  the  coke  burnt 
will  be  saved  by  using  a  feed-water  heater  which  reduces  the 


226  METALLURGICAL  CALCULATIONS. 

temperature  of  these  gases  to  200°  C.  and  sends  75  per  cent. 
of  the  heat  thus  abstracted  into  the  boiler  with  the   feed-water? 

7.92%. 
Problem  39. 

A  blast-furnace  gas  contains,  by  volume: 

Carbon  monoxide 23  per  cent. 

Carbon  dioxide 12 

Hydrogen 2 

Methane 2 

Water  vapor '. 3 

Nitrogen .  .58 

Its  temperature  is  20°  C.,  and  it  is  taken  to  a  gas  engine, 
mixed  with  the  theoretical  amount  of  dry  air  needed  for  perfect 
combustion,  and  compressed  to  4  atmospheres  tension  (total 
pressure)  before  being  ignited.  Neglect  the  heating  of  the 
gas-air  mixture  by  compression,  i.e.,  assume  the  temperature 
of  the  compressed  mixture  20°  C.  before  ignited.  Assume  that 
at  the  instant  of  ignition  the  volume  occupied  by  the  gases 
remains  constant,  so  that  the  specific  heat  at  constant  volume 
applies.  (Specific  heat  of  1  cubic  meter  at  constant  volume 
equals  specific  heat  at  constant  pressure — 0.09.) 

Required:  (1)  The  calorific  power  of  the  gas  per  cubic  meter, 
at  constant  pressure,  and  at  constant  volume.  928.5;  925.5. 

(2)  The  theoretical  temperature  at  the  moment  after  ignition 
has  taken  place.  1592°  C. 

(3)  The  maximum  pressure  exerted.  22§  atmospheres. 

(4)  The  volume  of  gas  needed  per  horse-power  hour,  at  a 
nechanico-thermal  efficiency  of  30  per  cent.       2.81  m3,  at  20°  C. 

Problem  40. 

Bituminous  coal  containing  carbon  78  per  cent.,  hydrogen 
5,  oxygen  8,  ash  8,  water  1,  is  used  in  a  gas  producer.  Assume 
the  calorific  power  of  the  coal  (water  formed  uncondensed)  as 
7480  Calories;  ashes  formed  12  per  cent.  Gas  formed  leaves 
producer  at  600°  C. ;  composition : 

Carbon  monoxide 35  per  cent. 

Carbon  dioxide 5 

Methane 5 

Hydrogen 5 

Nitrogen 50         " 


APPENDIX.  227 

Required:  (1)  The"vblume  of  gas  obtained  per  kilo,  of  coal 
burnt.  3.045  m3 

(2)  The  calorific  power  of  the  gas  per  cubic  meter. 

1633  Calories. 

(3)  The  proportion  of  the  calorific  power  of  the  fuel  obtain- 
able on  burning  the  gas.  66.5  per  cent. 

(4)  The  proportion  of  the  calorific  power  of  the  fuel  which 
has  been  sacrificed  in  making  the  gas,  assuming  it  burnt  cold. 

33.5  per  cent. 
Problem  41. 
Producer  gas  of  the  following  composition: 

Carbon  monoxide 28  per  cent. 

Carbon  dioxide 4 

Hydrogen 4 

Methane 2         " 

Water  vapor 1         " 

Nitrogen 61 

is  burned  with   10  per  cent,  more  air  than  theoretically  re- 
quired, both 'air  and  gas  being  preheated  to  1000°  C. 

Required:  The  theoretical  maximum  temperature  of  the  flame. 

2100°  C. 
Problem  42. 

Kiln -dried  peat  from  Livonia  contained  by  analysis: 

Carbon 49.70  per  cent. 

Hydrogen 5.33 

Oxygen ' 30.76 

Nitrogen 1.01 

Ash 13.23 

The  wet  peat,  as  taken  from  the  ground,  carried  75  per  cent. 
of  water,  when  air-dried  20  per  cent.,  and  when  kiln -dried 
none,  as  per  analysis. 

The  air-dried  peat  is  dried  in  a  kiln  by  means  of  a  current  of 
hot  air,  which  enters  the  kiln,  and  comes  in  contact  with  the 
freshly-charged  peat,  at  a  temperature  of  150°  C.,  while  it 
leaves  the  kiln,  near  the  discharge  end,  at  50°  C.  Outside  tem- 
perature 0°  C.,  outside  air  dry.  The  kiln  loses  by  radiation 
10  per  cent,  of  the  sensible  heat  of  the  hot  air  coming  into  it, 
the  rest  represents  the  sensible  heat  of  the  warm  moist  air  and 


228  METALLURGICAL  CALCULATIONS. 

dried  peat,  issuing  at  50°  C.,  and  the  heat  necessary  to  evaporate 
the  moisture.  The  air  required  is  heated  in  a  stove  where 
kiln-dried  peat  is  burned,  75  per  cent,  of  the  heat  generated 
being  transferred  to  the  air.  Mean  specific  heat  of  kiln-dried 
peat  0.25. 

Required:  (1)  The  practical  calorific  powers  of  wet  peat,  air 
dried  peat  and  kiln  dried  peat.  607;  3278;  4249. 

(2)  The  volume  of  air,  at  standard  conditions,  needed  for 
drying  one  metric  ton  of  air-dried  peat.  5,122  m3. 

(3)  The  percentage  degree  of  saturation,  with  moisture,  of 
the  issuing  air.  35.4. 

(4)  The  amount  of  kiln-dried  peat  required  to  be  burned  in 
the  stove  per  ton  of  air-dried  peat  put  through  the  kiln,  and  the 
percentage  of  the  total  fuel  necessary  for  this  purpose. 

70.5  kg.  9.25%. 
Problem  43. 

A  set  of  four  coke  ovens  produce  10,000  cubic  feet  of  gas 
per  hour,  measured  at  60°  F.,  and  having  the  composition  H2 
64.3  per  cent.,  CO  20.7,  CH<  5.4,  C2H<  0.5,  C6H«  0.5,  CO2  2.0, 
O2  1.0,  N2  5.6  per  cent.  Temperature  leaving  the  ovens  2900°  F. 
For  half  of  each  hour  the  whole  gas  is  burned  by  the  theoretical 
amount  of  cold  air,  as  it  passes  into  recuperators,  whence  the 
products  at  1900°  F.  pass  under  steam  boilers  where  their 
temperature  is  reduced  to  500°  before  passing  to  the  stack. 
For  the  second  half  of  each  hour  the  gases  pass  through  the 
recuperators  unmixed  with  air,  are  there  heated  to  1900°  F., 
and  at  that  temperature  pass  under  boilers  where  they  meet 
with  the  theoretical  quantity  of  air  needed  for  combustion, 
are  burned,  and  the  products  pass  to  the  stack  at  500°  F, 

Required:  (1)  The  horse-power  of  the  boilers  during  the  first 
30  minutes,  calling  1  horse-power  the  ability  to  evaporate  34J 
pounds  of  water  per  hour  at  212°  F.,  and  assuming  the  boilers 
to  produce  steam  representing  50  per  cent,  of  all  the  heat  re- 
ceived by  them  and  generated  within  them  (net  efficiency  50 
per  cent.).  23.4  H.  P. 

(2)  The  same,  for  the  second  30  minutes.  57.0  H.  P. 

Problem  44. 

A  plant  of  by-product  coke  ovens  uses  bituminous  coal  con- 
taining 76  per  cent,  of  carbon,  and  having  a  calorific  power 


APPENDIX.  229 

of  9000,  produces  coke  containing  an  average  of  86  per  cent, 
of  carbon,  and  having  a  calorific  power  of  7000,.  while  the  by- 
product tar  produced  contains  20  per  cent,  of  carbon.  The 
coke  weighs  70  per  cent,  and  the  tar  5  per  cent,  of  the  weight 
of  the  coal  used.  The  average  analysis  of  the  gases  for  the 
month  of  January,  1904  (samples  dried  before  analysis)  was: 

Carbon  dioxide 3.00  per  cent. 

Oxygen 0.50 

Carbon  monoxide 5.10 

Marsh  gas,  CH4 35.00 

fC2H4 2.13 

Illuminants  {  C3H8 1.06 

[C6H8 1.06 

Hydrogen 40.00 

Nitrogen.  . '. 12.15 

Required:  (1)  The  volume  of  by-product  gases,  at  standard 
barometric  pressure  and  at  60°  F.,  produced  per  ton  (2000 
pounds)  of  coal  used.  16,294  cubic  feet. 

(2)  The  proportion  of  the  calorific  power  of  the  coal  repre- 
sented by  the  calorific  powers  of  the  coke  and  the  gases. 

54.4%;  27.4%. 

(3)  Using  half  the  gases  produced  in  gas-engines,  at  an  effi- 
ciency of  conversion  into  power  of  25  per  cent.,  how  many 
horse-power-hours  could  be  thus  generated  per  pound  of  coal 
coked?  0.22  H.  P.  hours. 

Problem  45. 

A  gas  producer  uses  coal  which  analyzed:  total  carbon  75.68 
per  cent.,  oxygen  12.70  per  cent.,  hydrogen  4.50  per  cent., 
ash  7.12  per  cent.  The  ashes  produced  contain  21.07  per  cent, 
of  unburnt  carbon.  The  gas  carried  30.4  grams  of  moisture 
to  each  cubic  meter  of  dried  gas  collected,  and  the  latter  shows 
on  analysis: 

Carbon  dioxide 5.7  per  cent. 

Carbon  monoxide 22.0 

Methane  (CH4) 2.6 

Ethylene  (C2H4) 0.6 

Hydrogen 10.5 

Oxygen 0.4 

Nitrogen 58.2 


230  METALLURGICAL  CALCULATIONS. 

A  steam  blower  furnishes  blast,  forcing  in  the  outside  air  which 
carries  15  grams  of  moisture  per  cubic  meter,  measured  dry 
at  20°  C. 

Required:  (1)  The  volume  of  gas  (dry)  produced  per  kilo, 
of  coal  used.  4.337  m3. 

(2)  The  weight  of  steam  used  by  the  blower  per  kilo,  of 
coal  used.  0.2691  kg. 

(3)  The  weight  and  volume  of  air  blown  in,  per  kilo,  of  steam 
used  by  the  blower.  15.5  kg. ;  12.95  m3. 

(4)  The  percentage  of  the  steam  blown  in  which  is  not  de- 
composed in  the  producer,  assuming  the  moisture  in  the  gas  tc 
represent   steam  used  and  not   decomposed    (assumption  not 
strictly  accurate).  49.3%. 

(5)  The   mechanical   efficiency   of  the   blower,   assuming  it 
uses  steam  at  4  atmospheres  effective  pressure,  and  produces 
10  centimeters  of  water  gauge  pressure  in  the  ash  pit  of  the 
producer.  4.67%. 

Problem  46. 

Assume  the  gas  of  Problem  42  to  be  produced  by  the  com- 
bustion of  1000  kilos,  of  coal  per  hour  in  the  producers,  and 
to  be  burned  in  a  furnace  with  such  excess  of  air  (carrying 
at  20°  C.  15  grams  of  moisture  per  cubic  meter  measured  dry) 
that  the  chimney  gas,  analyzed  dry,  contains: 

Carbon  dioxide 12.7  per  cent. 

Nitrogen 80.6 

Oxygen 6.7 

These  products  of  combustion  enter  the  chimney  at  500°  C. 
and  leave  it  at  350°  C.,  their  velocity  entering  at  the  base  is 
4  meters  per  second.  Outside  air  20°.  The  efficient  draft  is 
2.5  centimeters  of  water  gauge,  measured  at  the  base;  assume 
this  90  per  cent,  of  the  total  head  of  the  chimney. 

Required:  (1)  The  diameter  of  the  chimney,  assuming  it 
round,  and  its  height,  assuming  its  section  uniform. 

1.52  meters;  41.3  meters. 

(2)  The   work   done   by  the   chimney,   expressed   in   horse- 
power. 2.6 

(3)  The  energy  efficiency  of  the  chimney,  i.e.,  the  ratio  of  the 
mechanical  work  it  performs  to  the  mechanical  equivalent  of  the 
heat  which  it  receives.  0.079  per  cent. 


APPENDIX.  231 

Problem  47. 

Assume  the  chimney  of  Problem  46  to  be  built  of  fire-brick 
of  an  average  thickness  of  60  centimeters;  that  the  gases  passing 
through  per  hour  are  1586  cubic  meters  of  nitrogen  and  air, 
315  cubic  meters  of  carbon  dioxide  and  234  cubic  meters  of 
water  vapor;  the  temperature  at  the  base  is  500°  C.,  at  the  top 
350°;  the  velocity  of  the  hot  gases  at  the  base,  4  meters  per 
second;  diameter  1.52  meters,  height  41.3  meters.  Assume 
further  the  coefficient  of  transfer  of  heat  from  gas  to  brick 

and  brick  to  air  to  be     tt _  ..J;        (in  C.  G.  S.  units) ,  and  the 

do, DUD 

velocity  of  the  wind  outside  to  be  20  kilometers  per  hour. 

Required'.  The  coefficient  of  conductivity  of  the  fire-brick 
in  C.  G.  S.  units.  0.31 


PART  II. 
IRON  AND  STEEL. 


233 


[AFTER  I. 
BALANCE  SHEET  OF  THE  BLAST  FURNACE. 

As  the  most  important  factor  in  the  production  of  the  most 
important  metal,  the  blast  furnace  is  the  most  important  fur- 
nace or  piece  of  metallurgical  apparatus  in  the  world.  It  is, 
therefore,  proper  that  we  should  commence  a  series  of  articles 
on  the  application  of  metallurgical  principles  and  calculations 
to  the  metallurgy  of  iron,  by  a  discussion  of  the  blast  furnace; 
and  since  this  discussion,  to  be  complete,  must  include  a  wide 
range  of  topics,  we  will  commence  with  the  simplest,  viz.:  the 
balance  sheet  of  materials,  entering  and  leaving  the  furnace. 
Later  we  can  discuss  the  balance  sheet  of  heat  entering  in, 
developed  within  and  leaving  the  furnace,  the  reactions  taking 
place  in  the  furnace,  the  action  of  hot  and  of  dried  blast,  the 
calculation  of  the  proper  constituents  of  the  charge,  the  tem- 
peratures attained  before  the  tuyeres,  unused  combustible 
energy  of  the  gases,  efficiency  of  the  hot-blast  stoves,  and  other 
interesting  and  practically  valuable  factors  in  the  running  of 
the  furnace. 

The  blast  furnace  may  be  regarded  from  several  points  of 
view;  we  will  mention  two.  First,  it  may  be  regarded  as  a 
huge  gas  producer,  run  by  hot,  forced  blast,  in  which  the  in- 
combustible portions  of  the  contents  are  melted  down  (with  a 
little  unburnt  carbon)  to  liquid  metal  and  slag,  and  are  run  out 
beneath,  while  the  gaseous  products  pass  upwards  through  50 
to  100  feet  of  burden,  and  escape  above.  The  escaping  gases 
are  primarily  of  Nthe  composition  of  producer  gas,  with  some 
of  its  carbonous  oxide  changed  to  CO2  by  the  oxygen  abstracted 
from  the"  burden,  with  some  CO2  added  from  the  decomposition 
of  the  carbonates  of  the  charge,  and  with  the  usual  increment 
of  moisture  from  the  charge  and  volatile  matter  (if  any)  from 
the  distillation  of  the  fuel.  From  this  point  of  view,  the  blast 
furnace  is  a  huge  gas  producer,  giving  a  rather  inferior  quality 
of  combustible  gas  in  very  large  quantities,  and  incidentally 

235 


236  METALLURGICAL  CALCULATIONS. 

reducing  to  metal  and  slag  the  burden  of  iron  ore  and  flux 
(limestone)  which  is  put  in  with  the  fuel.  The  treatment  of 
the  furnace  as  a  metallurgical  problem  may  then  proceed  as 
the  discussion  of  a  gas  producer,  witli  the  composition  of  the 
gas  produced  somewhat  modified  by  the  amount  of  oxygen 
given  up  to  the  gas  by  the  reducible  portions  of  the  charge  of 
the  furnace. 

The  other  viewpoint  is  to  regard  the  furnace  as  primarily 
an  apparatus  for  deoxidizing  or  reducing  iron  ore,  for  which 
purpose  the  ore  is  charged  with  sufficient  carbonaceous  fuel 
to  do  two  things,  viz.:  to  abstract  all  the  oxygen  from  the 
reducible  metallic  oxides,  and  to  furnish  enough  heat,  or  high 
enough  temperature,  to  melt  down  to  superheated  liquids  the 
pig  iron  and  slag  (combinations  of  irreducible  metallic  oxides) 
formed.  In  this  view,  the  fuel  must  supply  the  reducing 
energy  and  the  melting-down  or  smelting  requirements;  the 
first  by  acting  upon  the  metallic  oxides  at  a  red  to  a  white  heat 
and  abstracting  their  oxygen;  the  second,  by  being  burned  at 
the  foot  of  the  furnace  by  hot  air  blast,  and  there  generating 
the  heat  and  higher  temperatures  necessary  for  the  smelting 
down  of  the  already  reduced  materials. 

MATERIALS  CHARGED  AND  DISCHARGED. 

The  materials  put  into  a  blast  furnace  may  all  be  classed 
under  four  heads: 
Fuel  .........  ] 

Iron  ore  .....  \  Charged  at  the  throat. 

Fluxes  .......  j 

Blast  .............     Blown  in  at  the  tuyeres. 

The  materials  discharged  from  the  furnace  may  be  classed 
under  four  heads  also: 

iron  ...... 


••     Tapped  from  the  crucible. 

I  Passing  out  at  the  top. 

We  will  discuss  the  resolution  of  each  of  the  four  materials 
charged  into  the  four  avenues  of  escape. 

FUEL. 

The  fuel  used  is  sometimes  charcoal,  but  in  the  great  ma- 
jority of  cases  coke,  with  perhaps  some  raw  bituminous  coal 


BALANCE  SHEET  OF  BLAST  FURNACE.  237 

or  anthracite  coal,  or  in  a  few  cases  all  raw  bituminous  coal. 
The  composition  of  these  fuels  consists  of  moisture,  volatile 
matter,  fixed  carbon,  sulphur  and  ash  consisting  of  silica,  lime, 
iron,  alumina,  alkalies,  etc. 

The  moisture  is  driven  off  near  the  top,  and  goes  into  the 
gases  as  moisture.  The  volatile  matter  is  expelled  near  to  the 
top;  almost  all  of  it  goes  unchanged  into  the  gases,  but  part 
of  the  hydrocarbons  thus  expelled  may  be  decomposed  and 
deposit  fixed  carbon  on  the  iron  oxides,  etc.,  surrounding  them. 
This  carbon,  however,  will  take  up  oxygen  from  the  charge 
lower  down  in  the  furnace,  and  thus  eventually  pass  into  the 
gases  as  CO  or  CO2.  We  can,  therefore,  assume  without  error 
that  all  the  volatile  constituents  of  the  fuel  pass  into  the  gases, 
but  cannot  be  certain  in  exactly  what  state  of  combination, 
except  as  regards  the  moisture.  It  will  be  quite  exact  if  we 
know  the  ultimate  composition  of  the  volatile  matters  of  the 
coal,  as  so  much  carbon,  hydrogen,  oxygen,  nitrogen,  sulphur, 
etc.,  to  charge  them  thus  entirely  to  the  gases. 

The  fixed  carbon  all  finds  its  way  ultimately  into  the  dust 
or  the  gases,  either  as  CO,  CO2,  CH4  or  HCN,  or  alkali  cyanides, 
excepting  the  amount  represented  by  the  carbon  in  the  pig 
iron.  Subtracting  the  carbon  in  the  pig  iron  from  the  total 
fixed  carbon  in  the  fuel,  the  difference  can  safely  be  put  down 
as  entering  the  gases  or  being  in  the  dust  carried  away  by  the 
gases. 

The  sulphur  in  the  fuel  has  a  more  varied  history.  When  it 
is  partly  present  in  the  form  of  iron  pyrites,  some  may  go  into 
the  gases  as  sulphur  vapor,  and  eventually  be  burned  to  SO2 
when  the  gases  are  burned;  another  part  may  be  oxidized  in 
the  furnace  itself  to  SO2,  and  as  such  appear  in  the  gases;  the 
rest,  along  with  organic  sulphur,  passes  either  into  the  slag 
or  the  pig  iron.  Sulphur  passing  into  the  slag  seems  to  do  so 
as  calcium  sulphide,  CaS,  formed  by  some  such  reaction  as 

CaO  +  C  +  FeS  =  CaS  +  CO  +  Fe 
or,  if  the  sulphur  was  present  in  the  fuel  as  gypsum, 

CaS04  +  4C  =  CaS  +  4CO 
The  amount  of  sulphur  going  into  the  iron  depends  really  upon 


238  METALLURGICAL  CALCULATIONS. 

the  opportunity  for  it  to  go  into  the  slag.  If  the  temperature 
of  the  furnace  at  the  tuyeres  is  very  high,  and  especially  if  the 
slag  is  low  in  silica,  sulphur  will  keep  out  of  the  iron  and  go 
into  the  slag,  to  the  extent  of  ten  or  twenty  times  as  much 
being  in  the  slag  as  in  the  iron ;  but  if  the  temperature  is  low 
and  the  slag  rich  in  silica  the  reverse  may  be  the  case.  A 
high  temperature  and  a  high  percentage  of  lime  in  the  slag 
are  the  blast  furnace  manager's  means  of  keeping  down  the 
sulphur  in  the  iron,  although  high  magnesia  or  high  alumina 
are  also  efficacious.  In  casting  up  the  balance  sheet  it  can  be 
assumed  that  when  using  coke  or  charcoal  all  the  sulphur  of 
the  fuel  goes  either  into  the  slag  or  the  iron,  and  knowing  from 
the  analysis  of  the  pig  iron  made  how  much  goes  into  it,  the 
rest  can  be  calculated  as  going  into  the  slag  as  CaS.  If  raw 
coal  is  used,  it  is  uncertain  how  much  sulphur  goes  into  the 
gases,  and  an  exact  analysis  of  either  the  slag  or  gases,  for  sul- 
phur, in  addition  to  that  of  the  pig  iron,  would  be  necessary 
to  fix  its  distribution. 

The  ash  of  the  fuel  counts  in  with  the  other  incombustible 
ingredients  of  the  charge.  Some  of  the  silica  in  it  may  be 
reduced  to  silicon,  and  some  of  the  CaO  to  Ca,  to  form  CaS; 
while  most  of  the  iron  will  pass  into  the  pig  iron.  It  is  in 
most  cases  uncertain  whether  the  silicon  in  the  pig  iron  comes 
at  all  from  the  fuel  ash,  so  it  is  usual  to  assume  it  as  coming 
from  the  silica  of  the  ore  only;  as  to  the  iron,  it  is  best  to  assume 
it  all  reduced  to  the  metallic  state,  as  is  probably  always  the 
case. 

Besides  all  these  avenues  of  escape  for  the  constituents  of 
the  fuel,  it  is  sometimes  necessary  to  take  into  account  the 
possibility  of  some  of  it,  in  fine  particles,  being  carried  out  of 
the  furnace  bodily  with  the  outgoing  gases.  If  the  amount 
of  this  in  the  dust  is  determined,  it  must  be  subtracted  in  toto 
from  the  fuel  charged,  and  then  the  remainder  distributed  as 
just  discussed. 

Illustration. — A  blast  furnace  is  charged,  per  1,000  kilos,  of 
pig  iron  produced,  with  925  kilos,  of  coke,  containing  by  analysis: 
Fixed  carbon,  86  per  cent.;  volatile  carbon,  2;  hydrogen,  1; 
oxygen,  0.5;  nitrogen,  0.5;  sulphur,  1.0;  iron,  2;  silica,  5;  lime, 
1;  moisture,  1.  The  pig  iron  contains  3.5  per  cent,  of  carbon 
and  0.1  per  cent,  of  sulphur.  The  dust  carries  15  kilos,  of  dry 


BALANCE  SHEET  OF  BLAST  FURNACE.  239 

coke  per  metric  ton  of  pig  iron.     Required  the  distribution  of 
the  coke  in  the  furnace  per  ton  of  pig  iron  made: 

Charges.  Pig  Iron.     Slag.         Gases.  Dust. 


Coke  15.0 


VAJKC  

.   .    .  V^<J.\J 

^&- 

Dust  

...    15.0 

« 

.... 

Fixed  C.. 

...782.6 

"  C      35.0       ....     C 

747.6 

Volatile  C. 

...   18.2 

"       C 

18.2 

H        

.  ..     9.1 

"       H 

9.1 

O  

.  ..     4.5 

"       O 

4.5 

N 

;    .     45 

N 

4.5 

S 

.  ..     9.1 

"  S       1.0  S         8.1 

Fe   

.  .  .   18.2 

"  Fe    18.2       

SiO2  

.  ..   45.5 

....     SiO2  45.5 

.... 

CaO  

.  ..     9.1 

"       ....     CaO    9.1 

.... 

H2O.. 

9.3 

H2 

0      9.3 

ORE. 

Whatever  the  varieties  of  ore  used  they  can  be  averaged  to- 
gether, so  as  to  get  the  average  composition  of  the  ore  charged. 
Then,  knowing  its  weight  per  unit  of  pig  iron  made,  the  dis- 
tribution into  pig  iron,  slag,  gases  and  dust  can  be  made. 

There  may,  first  of  all,  be  blown  out  as  ore  dust  up  to  25  per 
cent,  of  the  ore  charged.  This  must  be  first  deducted  as  dry 
ore  and  then  the  rest  distributed  to  pig  iron,  slag  and  gases. 

The  moisture  of  the  ore,  also  any  carbonic  acid,  may  be  con- 
sidered as  going  over  bodily  into  the  gases.  The  sulphur  may 
partly  go  into  the  gases  if  present  as  iron  pyrites  (this  amount 
would  have  to  be  checked  by  an  analysis  of  the  gases  for  sul- 
phur or  hydrogen  sulphide),  but  mostly  into  the  slag  as  CaS. 
Some  of  it  may  be  put  down  as  going  into  the  pig  iron,  if  the 
sulphur  in  the  fuel  does  not  account  for  all  that  appears  in  the 
analysis  of  the  iron.  The  iron  oxides  present  must  be  as- 
sumed reduced  to  metallic  iron  sufficient  to  furnish  the  iron  in 
the  pig  iron  from  its  analysis;  the  excess,  if  any,  if  put  down 
as  going  into  the  slag  as  FeO;  all  the  oxygen  given  off  (the 
difference  between  the  weight  of  iron  oxide  in  the  ore  and 
the  sum  of  iron  going  in  the  pig  iron  and  ferrous  oxide  passing 
in  the  slag)  goes  to  the  gases.  If  there  is  not  enough  iron  in 
the  ore  to  account  for  all  in  the  pig  iron,  then  none  is  assumed 
to  go  into  the  slag. 


240  METALLURGICAL  CALCULATIONS: 

The  manganese  oxides  in  the  ore  furnish  the  manganese  in 
the  pig  iron,  the  excess  going  into  the  slag  as  manganous  oxide 
MnO,  the  oxygen  (by  difference)  goes  to  the  gases.  The  pro- 
portion of  manganese  reduced  to  metal  increases  with  the  tem- 
perature at  which  the  furnace  is  run  and  as  the  slag  is  less 
siliceous.  The  amount  reduced  is  known,  however,  only  by 
the  analysis  of  the  pig  iron. 

Zinc  in  the  ore  is  partly  found  as  ZnO  in  the  slag,  and  partly 
as  flakes  of  white  zinc  oxide  in  the  gases,  which  latter  partly 
deposit  in  the  dust  catcher  and  are  partly  carried  by  the  cur- 
rent of  gases  into  the  stoves  and  under  the  boilers.  The  rela- 
tive amounts  going  into  slag  and  gases  can  be  best  controlled 
by  analysis  of  the  slag. 

Copper,  silver,  gold,  nickel,  cobalt,  phosphorous,  antimony  and 
arsenic  are  almost  completely  reduced  into  the  pig  iron ;  careful 
analysis  of  the  latter  will  show  exactly  to  what  extent,  but 
without  this  careful  analysis  they  may  be  assumed  to  pass 
completely  into  the  iron.  Lead  is  mostly  carried  out  as  fume,  a 
small  amount  passes  into  the  pig  iron,  and,  if  present  in  quan 
tity,  a  large  amount  may  collect  as  metallic  lead  beneath  the 
pig  iron  and,  if  it  can,  soak  into  the  foundation  of  the  furnace. 

Alumina  usually  passes  completely,  as  such,  into  the  slag. 
When  present  in  large  amount,  producing  a  slag  rich  in  alu- 
mina, and  with  very  hot  blast,  the  pig  iron  may  contain  as 
much  as  1  per  cent,  of  aluminium,  the  oxygen  thereof  passing 
into  the  gases.  Magnesia  may  be  assumed  as  passing  com- 
pletely into  the  slag;  none  is  reduced.  Lime  goes  into  the  slag, 
except  a  not-unimportant  quantity  which  is  reduced  by  carbon 
in  the  presence  of  sulphur  compounds,  and  forms  CaS,  its 
oxygen  passing  into  the  gases ;  a  very  small  amount  may  go  as 
calcium  into  the  pig  iron.  Alkaline  metals  partly  go  into  the 
slag,  while  some  may  pass  into  the  gases  as  alkaline  cyanides. 
Titanium  oxide,  tungsten  oxide,  chromium  oxide  and  the 
oxides  of  molybdenum,  uranium,  vanadium  are  sometimes  re- 
duced in  small  amounts,  the  more  the  hotter  the  furnace  is  run 
and  the  more  basic  the  slag,  while  the  bulk  of  them  passes  into 
the  slag  as  the  lowest  oxide  which  each  is  capable  of  forming. 

Silica  mostly  goes  into  the  slag  as  SiO2,  but  a  portion  is  al- 
ways reduced  to  silicon  in  the  pig  iron.  The  amount  reduced 
is  greater  the  hotter  the  furnace  is  run,  the  more  slowly  it  is 


BALANCE  SHEET  OF  BLAST  FURNACE.  241 

run,  and  the  more  siliceous  the  slag.  In  some  cases  as  much 
as  one-quarter  of  all  the  silica  going  into  a  furnace  is  reduced 
to  silicon.  It  is  probably  reduced  only  by  carbon  dissolved  in 
iron  at  the  lower  part  of  the  furnace.  The  oxygen  of  the 
silica  reduced  goes  into  the  gases. 

Illustration. — 1956.8  kilograms  of  ore  is  charged  into  a  fur- 
nace per  metric  ton  of  pig  iron  made.  The  ore  analyses:  Fe203, 
71.43  per  cent.;  SiO2,  14.24;  CaO,  2.05;  MgO,  1.51;  MnO2,  4.15; 
SO3,  1.40;  H2O,  5.00;  Cu2O,  0.22  per  cent.  The  pig  iron  con- 
tains 93.03  per  cent,  iron,  3.27  carbon,  1.20  manganese,  0.08 
sulphur,  0.40  copper,  2.02  silicon.  Assume  the  dry  ore  dust  to 
weigh  4  per  cent,  of  the  weight  of  ore  charged,  and  that  there 
is  no  sulphur  found  in  the  gases,  but  all  the  sulphur  in  the  pig 
iron  comes  from  the  fuel.  Cast  up  the  distribution  of  the  ore: 

Charge.  Pig  Iron.     Slag.          Gases.       Dust. 

Ore 1956.8kg. 


Dust 78.1  Ore   78.1 

Fe2O3 1339.2  "  Fe  930.3  FeO      9.1  O  499.8       .... 

SiO2 267.0  "  Si    20.2  SiO2  223.6  O     22.2       

MnO2 77.8  "  Mn  12.0  MnO    48.0  O     17.8       

Cu20 4.1  "  Cu     3.6       ....         0      0.5       .... 

CaO  38  4    "  '  Ca°  2(U 

Ca° 38'4  •;••      \Ca      13.10       5.2 

MgO 28.3    "         ....     MgO     28.3     

SO3 26.3    "         S          10.50       15.8     

H20 97.6    "         H20  97.6     .... 

In  calculating  the  above  we  note  that  the  dust  is  dry,  and 
weighs  4  per  cent,  of  the  ore,  making  78.1  kilos,  of  dry  dust, 
representing  82  kilos,  of  moist  ore,  leaving  1874.8  kilos,  of 
moist  ore  to  be  distributed,  plus  the  3.9  kilos,  of  water  from  the 
dust,  which  also  goes  into  the  gases.  This  1874.8  kilos,  contains 
the  weights  given,  calculating  from  its  analysis.  The  1339.2 
kilos,  of  Fe2O3  contains  937.3  kilos,  of  iron;  but  there  are  only 
930.3  kilos,  in  the  ton  of  pig  iron,  therefore  the  other  7.0  kilos. 

72 

must  go  into  the  slag  as  7.0X-^=  9.1  kilos,  of   ferrous  oxide, 

oo 

while    1339.2-  (930.3  +  9.1)  =  499.8,    the    weight    of    oxygen 


242  METALLURGICAL  CALCULATIONS. 

going  into  the  gases.  The  267.0  kilos,  of  silica  contains  124.6 
kilos,  of  silicon,  but  there  are  only  20.2  kilos,  in  the  pig  iron, 
therefore,  104.4  kilos,  must  remain  unreduced,  passing  into  the 

60 

slag  as  104.4  X  ™  =  223.6  kilos,   of  silica,  while  267-  (20.2  + 
Zo 

223.6)  =  22.2  kilos,  of  oxygen  goes  into  the  gases.  Another, 
and  equally  logical  procedure,  is  to  start  with  the  20.2  kilos. 
of  silicon  in  the  pig  iron,  which  must  have  required  20. 2  X 

60 

—  =  42.4  kilos,  of  silica  to  furnish  it,  yielding  42.4-  20.2  = 

Zo 

22.2  kilos,  of  oxygen  to  the  gases,  and  leaving  267.0—  42.4  = 
223.6  kilos,  of  silica  unreduced  to  go  into  the  slag. 

The  12.0  kilos,  of  manganese  in  the  pig  iron  would  be  reduced 

87 

from  12.0  XTT  =  19.0  kilos,  of  MnO2,  furnishing,  therefore,  7.0 
oo 

kilos,  of  oxygen  to  the  gases,  and  leaving  77.8—19.0  =  58.8 
kilos,  of  MnO2  to  go  into  the  slag  as  MnO.  The  molecular 
weights  of  MnO  and  MnO2  being  respectively  71  and  87,  there  is 

58.8  X^r  =  48.0  kilos,  of  MnO  going  into  the  slag,  while  10.8 
o7 

kilos,  more  of  oxygen  will  be  supplied  to  the  gases,  making  alto- 
gether 10.8  +  7  =  17.8  kilos,  of  oxygen  given  up  by  the  MnO2. 
The  4.1  kilos,  of  Cu2O  contain  3.6  kilos,  of  copper,  all  of  which 
enters  the  pig  iron  and  contributing  0.5  kilos,  of  oxygen  to  the 
slag. 

The  38.4  kilos,  of  CaO  must  supply  enough  Ca  to  form  CaS 

00 

with   the   S   of  the   SO3.     The  latter   quantity  is   26.3  X^n  = 

oU 

40 
10.5  kilos.,  to  supply  which  there  is  needed  10.5  X  ^5  =  13.1 

OA 

56 
kilos,  of  calcium.     The  latter  will  be  supplied  by  IS.lXTTj  — 

18.3  kilos,  of  CaO,  furnishing  5.2  kilos,  of  oxygen  to  the  gases, 
and  leaving  38.4—  18.3  =  20.1  kilos,  of  CaO  unreduced  to  go 
into  the  slag.     The  MgO  in  the  ore  goes  directly  into  the  slag. 
The  oxygen  of  the  SO3  goes  into  the  gases.     The  5  per  cent,  of 


BALANCE  SHEET  OF  BLAST  FURNACE.       243 

water  of  the  whole  quantity  of  ore  charged  goes  into  the  gases, 
as  vapor. 

FLUX. 

The  flux  is  used  for  the  purpose  of  making  a  fusible  slag 
with  the  slag-forming  ingredients,  contributed  by  the  ore  and 
fuel.  If  we  consider  the  distribution  of  ore  and  fuel  given  in 
the  preceding  illustrations  we  see  that  the  chief  material  to  be 
fluxed  is  silica,  with  smaller  quantities  of  FeO,  MnO,  CaO, 
MgO  and  CaS.  The  cheapest  and  most  available  material  to 
flux  silica  is  limestone,  the  slag  formed  being  a  silicate  of  lime, 
magnesia  and  alumina,  with  CaS  and  smaller  quantities  of 
other  basic  oxides.  We  will  not  discuss  at  present  the  con- 
siderations governing  the  amount  of  flux  used,  since  this  is  a 
calculation  requiring  separate  treatment,  as  the  proper  working 
of  the  furnace  depends  fundamentally  upon  it.  We  may  re- 
mark here  that  enough  flux  must  be  used  to  make  an  easily 
fusible  fluid  slag,  rich  enough  in  lime,  magnesia  or  alumina  to 
carry  away  satisfactorily  the  bulk  of  the  sulphur,  and  so  pro- 
duce good  pig  iron. 

The  flux  usually  contains  CaO,  MgO,  APO3,  SiO2,  FeO,  CO2 
and  H2O.  Its  H2O  and  CO2  are  driven  off  in  the  upper  third 
of  the  furnace,  and  may  be  put  down  as  going  as  such  into  the 
gases.  The  FeO  may  be  reduced  if  the  slag  is  very  clean,  but 
under  ordinary  conditions  may  be  put  down  as  all  going  into 
the  slag,  unless  in  quite  large  amount,  because  the  iron  in  ore 
and  fuel  usually  supplies  the  total  weight  of  iron  in  the  pig  iron. 
The  silica  and  alumina  may  be  carried  over  bodily  into  the 
slag.  The  magnesia  can  be  put  at  once  into  the  slag,  but  the 
lime  cannot  in  many  cases  be  treated  that  way,  because  quite 
frequently  some  is  needed  to  supply  calcium  for  the  sulphur 
of  the  fuel.  In  the  fuel  previously  illustrated,  for  instance, 
there  is  not  enough  CaO  present  to  furnish  Ca  for  the  S,  whence 
it  follows  that  some  CaO  from  the  flux  will  be  needed  to  make 
up  the  deficit.  We  may,  in  such  a  case,  either  consider  all  the 
CaO  of  the  fuel  to  form  CaS  with  part  of  the  sulphur,  and  then 
take  enough  CaO  from  the  flux  to  unite  with  the  remainder, 
or,  it  is  equally  permissible  to  take  all  the  CaO  necessary  to 
furnish  Ca  to  all  the  sulphur  of  the  fuel,  and  to  let  the  CaO 
of  the  fuel  figure  as  passing  entirely  into  the  slag.  The  latter 
requires  a  little  less  calculation. 


244  METALLURGICAL  CALCULATIONS. 

Illustration. — A  blast  furnace  receives  503  kilos,  of  lime- 
stone flux  per  metric  ton  of  pig  iron  made,  which  analyses  CaO, 
29.68  per  cent.;  MgO,  20.95;  SiO2,  3.07;  APO3,  2.66;  FeO,  0.48; 
CO2,  42.66;  H2O,  0.50  per  cent.  Assume  8.1  kilos,  of  sulphur 
in  the  fuel,  for  which  the  flux  must  provide  calcium.  Required 
the  distribution  of  the  flux,  assuming  it  to  make  no  dust: 

Charge.                  Pig  Iron.  Slag.  Gases. 

Flux 503.0kg 


135.1 

MgO 105.4  "  ....  MgO          105.4 

SiO2 15.4  "  ....  SiO2            15.4 

A1203 13.4  "  ....  APO3          13.4 

FeO 2.4  "  ....  FeO              2.4 

CO2 ....214.6  "  ....               CO2     214.6 

H20 2.5  "  ....                H2O        2.5 

The  only  calculation  needed  above  is  that  8.1  kilos,  of  sulphur 

40 

require  8.1  Xr«  =  10.1    kilos,    of    calcium,    which    would    be 

32 

furnished  by  lO.lX^Tj  =  14.2  kilos,   of  lime,  leaving  4.1  kilos. 

of  oxygen  to  go  into  the  gases  and  135.1  of  lime  to  go  into  the 
slag. 

BLAST. 

The  remaining  item  needed  to  complete  the  balance  sheet  is 
the  amount  of  blast.  This  may  be  roughly  estimated  by  ob- 
taining the  piston  displacement  of  the  blowing  engines,  and 
assuming  a  coefficient  of  delivery  into  the  furnace.  This  is 
very  rough,  because  the  efficiency  is  not  known,  and  may  vary 
anywhere  between  0.5  and  0.95.  Another  rough  approximation 
may  be  obtained  by  observing  the  pressure  of  the  blast,  its 
temperature,  the  back  pressure  in  the  furnace,  and  knowing 
the  area  of  the  tuyeres,  and  assuming  a  coefficient  of  contrac- 
tion of  the  hot  air  jet  as  it  emerges  from  the  tuyeres.  Here, 
again,  are  several  uncertain  factors,  and  the  coefficient  may 
vary  between  0.9  and  0.98.  Calculations  on  this  basis  are  very 
rough. 


BALANCE  SHEET  OF  BLAST  FURNACE.  24f 

\ 

The  only  satisfactory  way  to  determine  the  blast  is  to  care^ 
fully  analyze  the  gases,  determining  carefully  all  the  carbon, 
oxygen  and  nitrogen  which  they  contain.  Since  the  carbon 
comes  only  from  the  charges,  the  amount  of  gases  produced  per 
unit  of  pig  iron  made  becomes  known,  and  thence  the  oxygen 
and  nitrogen  contained  in  them.  These,  minus  the  oxygen  and 
nitrogen  coming  from  the  solid  charges,  leave  the  oxygen  and 
nitrogen  which  must  have  come  from  the  blast.  The  oxygen 

3 
in  the  blast,  minus  —  the  nitrogen,  gives  the  oxygen  entering 

as  water  vapor;  but  this  last  calculation  is  not  so  satisfactory 
as  to  observe  the  atmospheric  conditions,  and  calculate  the  air 
and  moisture  on  the  basis  of  the  contained  nitrogen. 

The  blast  contains  oxygen,  nitrogen  and  moisture.  All  its 
constituents  pass  into  the  gases,  being  put  down  as  so  much 
oxygen,  nitrogen  and  hydrogen.  Just  how  much  of  that  hydro- 
gen gets  into  the  gases  as  free  hydrogen  and  how  much  as 
water  vapor  is  not  known.  Argon  and  other  rare  gases  in  the 
blast  are  counted  and  treated  as  nitrogen.  The  carbonic  acid 
of  the  air  is  present  relatively  in  such  a  small  amount  that  it 
can  be  neglected,  as  far  as  all  ordinary  calculations  are  con- 
cerned. 

Problem  51. 

A  blast  furnace  at  Herrang,  Sweden,  is  run  on  ore  briquettes 
made  by  pressing  and  calcining  fine  concentrates.  The  analyses 
of  briquettes,  charcoal  and  limestone  flux  are  as  follows  (see 
Journal  Iron  and  Steel  Institute,  I.,  1904): 

Briquettes.       Limestone.        Charcoal. 

Fe2O3 85.93  0.18  0.32 

FeO 3.96  C  80.31 

SiO2 5.50  3.14  0.19 

MnO 0.63  N  0.08 

A12O3 0.76  0.32          O  3.54 

CaO 2.23  53.74  0.89 

MgO 0.97  0.17  0.10 

P2O5 0.006  0.006  0.0068 

S 0.010  0.001  0.017 

Cu 0.007     CO2  42. 42       H2O  14.04 

K2O  0.50 


246  METALLURGICAL  CALCULATIONS. 

The  pig  iron  contains  phosphorus,  0.012  per  cent.;  sulphur, 
0.007;  manganese,  0.025;  silicon,  0.60;  carbon,  2.70;  iron, 
96.656.  There  is  used  in  charging  the  furnace: 

Briquettes 1,190  pounds 

Limestone. 90 

Charcoal 530 

And  the  fuel  consumption  is  682  pounds  of  charcoal  per  1,000 
pounds  of  pig  iron  made. 

The  gases  at  the  throat  (dried)  analyze:  N2,  57.3  per  cent.; 
CO,  23.1;  CO2,  14.8;  H2,  4.3;  CH4,  0.5  (Rinman).  Assume 
blast  dry.  Dust  in  gases  neglected. 

Required:  (1)  A  balance  sheet  of  materials  entering  and  leav- 
ing the  furnace,  per  1,000  pounds  of  pig  iron  made.  (2)  The 
percentages  of  iron,  manganese,  silicon,  sulphur  and  phosphorus 
going  into  the  furnace,  which  go  into  the  pig  iron. 

Solution: — (1)     See  table  opposite  page. 

(2)  The  total  iron  in  the  charge  is  969.2  kilos.,  while  that  in 
the  pig  iron  is  966.6;  the  efficiency  of  the  reduction  of  iron  is 
therefore  99.7  per  cent. 

The  total  manganese  in  the  charge  is  9.6X=y  =  7.4  kilos., 

of  which  only  0.25  gets  into  the  pig  iron,  or  3.4  per  cent. 

The  total  silica  charged  is  89.1  kilos.,  representing  41.6  kilos. 
of  silicon,  of  which  6.0  kilos,  enters  the  pig  iron,  or  14.4  per 
cent. 

The  sulphur  charged  is  0.270  kilos.,  of  which  the  pig  iron 
contains  0.07,  or  25.9  per  cent. 

The  phosphorus  charged  is  0.063  kilos.,  while  the  analysis  of 
the  pig  iron  shows  in  it  0.12  kilos.  It  is  thus  evident  that  all 
the  phosphorus  goes  into  the  pig  iron;  for  while  the  analysis 
shows  more  phosphorus  in  the  pig  iron  than  was  put  into  the 
furnace,  yet  the  divergence  is  evidently  due  to  segregation  or 
concentration  of  phosphorus  in  the  sample  taken,  and  the  prac- 
tical conclusion  is  that  all  the  phosphorus  in  the  charge  finds  its 
way  into  the  pig  iron. 

NOTES  ON  THE  BALANCE  SHEET. 

The  Fe2O3  of  the  ore  is  assumed  all  reduced,  because  the 
920.4  kilos,  of  iron  in  it  is  less  than  the  966.6  kilos,  of  iron 
known  to  be  in  the  1,000  kilos,  of  pig  iron  from  its  analysis. 


BALANCE  SHEET  OF  BLAST  FURNACE.  247 


Solution.-    (1) 

BALANCE  SHEET  (PER  1000  OF  PIG  IRON). 
Charges  Pig  Iron  Slag  Gases 


Ore 
Fe2O3 

1530.2 
1314  9 

Fe         920  .4 

o 

394.5 

FeO 
SiO2 
MnO 
A12O3 

60.6 
84.2 
9.6 
11  6 

Fe           46.2       FeO 
Si              6.0       SiO2 
Ma           0.25     MnO 
....         A12O3 

1.2 
69.6 
9.3 
11.6 

0 
O 
O 

13.2 
8.6 
0.1 

CaO 

34  1 

CaO 

34  1 

o 

0.03 

MgO 

14.8 

MgO 

14.8 

p2Q5 

0  092 

P              0  .  04 

O 

0.05 

S 
Cu 

0.153 
0  11 

S               0.07     CaS 
Cu            0  11 

0.19 

o 

0  01 

Limestone 

115.8 

Fe2O3 

0  2 

FeO 

0.2 

O 

0.02 

SiO2 

3  6 

.    .    .         SiO2 

3.6 

A12O3 

0  4 

A12O3 

0  4 

CaO 

62.2 

CaO 

62.2 

MgO 
P2O5 

0.2 

0  007 

MgO 
P              0  003 

0.2 

o 

0.00 

s 

0  001 

CaS 

0.0 

CO2 

49  1 

CO2 

49  1 

Charcoal 

682.0 

c 

547  7 

C             27  .  0 

c 

520.7 

N 

0  5 

N 

0.5 

o 

24  1 

o 

24  1 

Fe203 
SiO2 

2.2 
1  3 

FeO 

.    .    .          SiO2 

2.0 
1.3 

O 

0.2 

CaO 

6  1 

CaO 

5  9 

o 

0  06 

MgO 

0  7 

MgO 

0.7 

p2Q5 

0  046 

P              0  .  02 

o 

0.03 

s 

0  116 

CaS 

0.25 

K2O 

3.4 

K2O 

3.4 

H2O 

95  8 

H2O 

95  8 

Blast 

2416.8 

O2 

557  7 

o 

557  7 

N2 

1859  1 

N2 

1859  1 

Totals  4744.0  1000.0  220.8  3543.7 


248  METALLURGICAL  CALCULATIONS. 

The  FeO,  however,  cannot  be  assumed  all  reduced,  because  it 
would  furnish  47.1  kilos,  of  iron,  and  there  is  only  966.6— 
920.4  =  46.2  kilos,  of  iron  yet  to  be  supplied.  We,  therefore, 
put  down  46.2  kilos,  of  iron  as  going  to  the  pig  iron,  thus  fur- 
nishing all  the  iron  in  the  pig  iron,  and  leaving  0.9  kilos,  of 
iron  to  go  over  into  the  slag  as  1.2  kilos,  of  FeO.  Having  thus 
allowed  for  all  the  iron  in  the  pig  iron,  the  Fe2O3  in  the  lime- 
stone and  fuel  must  be  assumed  as  passing  entirely  into  the 
slag  as  FeO. 

The  6  kilos,  of  silicon  in  the  pig  iron  is  put  down  as  coming 
entirely  from  the  SiO2  of  the  ore,  of  which  14.6  kilos,  is  thus 
used  up,  leaving  15.6  kilos,  to  go  into  the  slag.  The  SiO2  of 
'flux  and  fuel  must  then  be  regarded  as  passing  entirely  into 
the  slag. 

The  0.25  of  manganese  in  the  pig  iron  comes  from  the  MnO 
of  the  ore,  requiring  0.35  of  MnO,  and  leaving  9.3  of  MnO  to 
go  into  the  slag. 

The  A12O3  and  MgO  of  ore,  flux  and  fuel  go  bodily  into  the 
slag. 

The  sulphur  in  the  ore,  0.153  kilos.,  is  more  than  enough  to 
supply  the  0.07  kilos  in  the  pig  iron.  We,  therefore,  put  down 
0.07  kilos,  as  going  into  the  pig  iron,  supplying  all  the  latter 
contains,  and  calculate  the  remaining  0.083  kilos,  to  CaS  going 
into  the  slag.  The  CaO  necessary  to  furnish  this  calcium  is 
56  for  every  32  of  sulphur  (CaO  =  56,  S  =  32),  or  0.14  kilos., 
which,  therefore,  must  be  deducted  from  the  34.1  kilos,  of  CaO 
present  in  the  ore.  The  oxygen  of  this  0.14  kilos,  of  CaO  finds 
its  way  into  the  gases. 

The  0.092  kilos,  of  P205  present  in  the* ore  contain  only  0.04 
kilos,  of  phosphorus,  and  since  the  pig  iron  contains,  from  its 
analysis,  0.12  kilos.,  we  may  assume  all  of  this  going  into  the 
pig  iron.  The  same  remarks  are  true  of  the  P2O5  in  flux  and 
fuel;  altogether,  they  come  somewhat  short  of  supplying  all 
the  phosphorus  in  the  pig  iron,  and  are,  therefore,  considered  as 
completely  reduced.  The  copper  goes  entirely  into  the  pig  iron, 
although  not  given  in  the  analysis. 

The  Fe2O3  of  the  limestone  must  be  transferred  entirely  as 
FeO  to  the  slag,  since  all  the  iron  needed  for  the  pig  iron  has 
been  already  provided.  The  same  is  true  of  the  Fe203  of  the 
fuel;  and  an  analogous  statement  applies  to  the  SiO2  and  sul- 


BALANCE  SHEET  OF  BLAST  FURNACE  249 

phur  of  both  flux  and  fuel.  The  sulphur  of  the  fuel  does  not 
produce  an  amount  of  CaS  which  counts  in  significant  figures, 
and  the  CaO  required  is  likewise  insignificant,  as  is  also  the 
oxygen  thus  furnished  the  gases.  In  such  cases,  instead  of 
ignoring  the  item  altogether,  or  putting  down  wholly  insignifi- 
cant quantities,  the  amounts  are  expressed  as  0  00,  denoting  no 
significant  amount. 

The  fixed  carbon  of  the  fuel,  only,  furnishes  the  carbon  in 
the  pig  iron,  the  rest  going  into  the  gases.  The  blast  is  calcu- 
lated as  follows: 

12 

Carbon  in  CO2  of  flux  =  49.1X-7T  =       13.39  kilos. 

44 

Carbon  in  gases  from  fuel  =    520 . 70     " 

Carbon  in  gases  altogether  -    534.09     " 

Carbon  in  1  cu.  meter  of  gas  (0.231 +  0.148 -f 

0.005)  X 0.54  =0.20736     " 

Volume  of  gas  per  1,000  of  pig  iron  534.09  -r- 

0.20736  =    2575.6    m3 

Nitrogen  in  this  gas  2575.6X0.573  =     1475.9     " 

Weight  of  nitrogen  1475.9X  1.26.  =     1859.6  kg. 

Nitrogen  from  fuel  =          0 . 5    4 

Nitrogen  from  blast =     1859. 1    " 

Oxygen  from  blast  1859.1  X0.3 -      557.7    " 


CHAPTER  II. 

CALCULATION  OF  THE  CHARGE  OF  THE  BLAST 
FURNACE. 

In  the  last  chapter  we  discussed  the  balance  sheet  of  ma- 
terials entering  and  leaving  the  furnace,  showing  the  distribu- 
tion of  the  ingredients  of  the  charge  and  the  blast  into  the 
various  products  and  by-products  of  the  furnace.  We  did  not 
there  go  into  the  question  as  to  how  the  proportions  of  the 
charge  are  determined  by  the  metallurgist  in  charge  of  the 
furnace.  There  are,  however,  very  few  factors  of  the  charge 
which  can  be  controlled  at  will.  The  blast  furnace  reduces 
practically  all  of  the  iron  present  in  the  ore  into  the  pig  iron, 
and,  therefore,  if  the  ore  contains  50  per  cent,  of  metallic  iron 
and  the  pig  iron  90  per  cent.,  it  will  take  0.90 -i- 0.50  =  180 
parts  of  ore  to  furnish  the  iron  in  100  parts  of  pig  iron.  The 
amount  of  ore  to  be  used  per  unit  of  pig  iron  made  is  there- 
fore fixed  by  the  richness  or  poverty  of  the  ore,  and  is  not 
capable  of  variation.  The  amount  of  fuel  used  per  unit  of 
pig  iron  made  is  not  fixed  a  priori,  as  is  the  amount  of  ore, 
but  is  governed  by  the  calorific  requirements  of  the  furnace 
while  in  operation.  If  the  pig  iron  and  slag  run  colder  than 
they  should,  it  is  evident  that  more  heat  must  be  put  in  or 
developed  within  the  furnace,  which  the  manager  promptly 
proceeds  to  accomplish  by  increasing  the  temperature  of  the 
air  blast  (if  he  can),  or  by  relatively  increasing  the  amount  of 
fuel  in  the  charge  (which  he  always  can).  The  ratio  of  the 
weight  of  the  ore  and  flux  in  the  charge  to  the  weight  of  fuel 
used  is  called  the  burden  of  the  furnace,  and  in  practice  it  is 
usual  to  charge  at  one  time  a  fixed  weight  of  fuel,  and  to  vary 
the  burden  according  to  the  heat  requirements  of  the  furnace. 
Changing  the  burden  is  therefore  only  another  expression  for 
changing  the  relative  amount  of  fuel  used,  and  this  is  varied 
simply  from  observation  of  the  temperatures  of  iron  and  slag 
and  the  conclusions  therefrom  as  to  whether  the  burden  is  too 

250 


CALCULATION  OF  THE  CHARGE.  251 

heavy  or  unnecessarily  light.  The  amount  of  blast  used  is 
another  factor  relatively  fixed  per  unit  of  pig  iron  produced. 
Blow  more  blast,  and  more  pig  iron  is  produced;  blow  no  blast 
(bank  the  furnace),  and  no  pig  iron  is  made;  the  ratio  is  not 
quite  exact,  but  it  is  quite  nearly  true  that,  other  conditions 
being  equal,  the  output  of  pig  iron  is  nearly  proportional  to  the 
amount  of  blast  blown.  Variations  in  the  temperatures  of  the 
blast  produce  important  changes,  which  will  be  separately  dis- 
cussed. 

The  amount  of  flux  used  is  really  the  one  factor  in  which  the 
manager  has  the  greatest  freedom  of  action.  The  amount  of 
this  indispensable  substance  used  is  determined  by  many  fac- 
tors, and  can  be  varied  between  quite  wide  limits  without 
fundamentally  deranging  the  furnace.  It  is  here  a  question  of 
using  sufficient  flux  in  the  charge  to  make  with  the  unreduced 
constituents  of  ore  and  ash  of  the  fuel  a  slag  which  shall  be 
well-fused  at  the  temperature  of  the  furnace,  shall  carry  off 
considerable  sulphur,  if  much  is  present,  and  shall  not  corrode 
the  lining  of  the  furnace.  These  considerations  are  so  im- 
portant, and  often  so  little  understood,  that  we  will  discuss 
them  more  at  length. 

CALCULATION  OF  THE  FLUX  AND  SLAG. 

From  the  balance  sheet  of  problem  51  it  will  be  seen  that  the 
metallurgist  running  the  Swedish  blast  furnace  at  Herrang, 
used,  per  1,000  of  pig  iron  made,  1530.2  of  ore,  115.8  of  lime- 
stone flux,  682  of  charcoal,  and  blew  in  2416.8  parts  by  weight 
of  blast.  It  may  with  safety  be  believed  that  the  amount  of 
charcoal  used  was  the  minimum  which  he  found  by  experience 
necessary  to  keep  his  furnace  at  proper  temperature;  and  most 
American  blast  furnace  managers  will  wonder  how  he  could 
get  along  with  so  little.  The  amount  of  ore  used  was  the 
necessary  proportion  to  furnish  the  iron.  The  amount  of  blast 
wa?  probably  all  that  could  be  gotten  out  of  the  blowing  ap- 
paratus with  which  the  furnace  was  provided.  Finally,  the 
amount  of  flux  was  that  amount  necessary  to  make  a  proper 
slag.  Let  us  investigate  the  question  as  to  how  its  amount  was 
determined  and  the  characteristics  of  the  proper  slag. 

The  ingredients  of  the  slag  produced  in  the  case  in  question 
are,  from  the  balance  sheet: 


252  METALLURGICAL  CALCULATIONS. 

From  Ore.  From  Flux.  From  Fuel.     Total 

SiO2 69.6  3.6  1.3  74.5 

APO3 11.6  0.4  ....  12.0 

CaO 34.0  62.2  5.9  102.1 

MgO 14.8  0.2  0.7  15.7 

FeO 1.2  0.2  2.0  3.4 

MnO 9.3  9.3 

K2O ....  3.4  3.4 

CaS..                         0.2  0.25  0.45 


140.7  66.6  13.55        220.85 

And  the  percentage  composition  of  the  slag: 

SiO2      =  33.73  per  cent.         FeO    =  1.54  per  cent. 
APO3     =     5.43       "  MnO  =  4.21       " 

CaO      =26.23       "  K2O    =  1.54       " 

MgO     =7.11       "  CaS     =0.20       " 

The  crucial  question  now  presents  itself,  "  What  guided  the 
metallurgist  in  choosing  the  quantity  of  lime  stone  used,  and 
in  making  a  slag  of  the  above  composition?  "  The  answer  to 
this  will  develop  the  whole  practice  of  fluxing. 

Primarily,  the  fundamental  guide  in  this  matter  is  previous 
experience,  as  revealed  in  recorded  analyses  of  slags  which 
have  worked  properly.  Such  analyses  have  been  made  for  a 
hundred  years,  and  freely  published;  they  show  what  com- 
positions of  slag  have  been  found  practicable  and  suitable  in 
blast  furnace  practice.  As  reported  by  the  chemists,  the  analyses 
show  the  varying  percentages  of  SiO2,  APO3,  CaO,  MgO,  FeO, 
MnO,  etc.,  found  in  actual  slags  made  in  successful  practice, 
and  this  information  would  be  the  metallurgist's  guide  in  calcu- 
lating the  amount  of  flux  to  use  to  produce  a  good  slag.  On 
studying  these  analyses,  we  find  that  silica  and  lime  are  the  pre- 
dominating constituents  of  all  blast  furnace  slags,  the  reason 
being  that  silica  is  the  principal  material  to  be  fluxed,  and  that 
lime  (from  limestone)  is  the  cheapest  material  which  will  flux 
it,  and  form  a  fusible  slag. 

Analyses  of  numerous  blast  furnace  slags  show  the  following 
variations  of  composition: 

SiO2      25  to  65  per  cent.  FeO         0  to    6  per  cent. 

APO3      3  "  30       "  MnO        0  "  14       " 

TiO2       0  "  10       "  K2O 


CaO      12  "  50       "  Na2O  /     °  ' 

MgO       0  "  18       "  CaS          0  "     9 


CALCULATION  OF  THE  CHARGE.  253 

The  above  limits  are  not  reached  simultaneously  by  one  and 
the  same  slag.  The  ordinary  variations  may  be  summarized 
as  follows  (according  to  Ledebur) : 

SiO2  APO3  CaO  +  MgO 

Producing  gray  iron,  using  char- 
coal    45—65  10—  5  45—25 

Producing  gray  iron,  using  coke.   30—35  15 — 10  50 — 55 

Producing  white  iron,  using  char- 
coal    45—50  10—  5  45— 

Producing  white  iron,  using  coke  30 — 40  10 —  5  60 — 55 

Producing  spiegeleisen,  using 

coke 30  10  55—45 

In  the  latter  case  there  is  present  5 — 15  per  cent,  of  MnO. 

Among  these  varied  compositions,  however,  there  are  various 
degrees  of  fusibility,  and  of  suitability  to  the  blast  furnace's 
needs.  The  most  easily  fusible  slags  are  those  (considering 
only  the  main  ingredients)  which  contain  35  per  cent,  of  lime, 
and  in  which,  if  alumina  is  present,  each  1  per  cent,  of  alumina 
is  balanced  by  the  presence  of  0.5  per  cent,  additional  lime. 
This  rule  gives  good  slags  up  to  about  65  per  cent,  of  lime  and 
alumina  counted  together.  Another  observation  is,  that  with 
33  to  40  per  cent,  of  lime  in  the  slag,  the  amount  of  silica  and 
alumina  together  being  some  60  to  70  per  cent.,  it  is  possible  to 
change  the  relations  of  silica  to  alumina  within  large  limits 
(i.e.,  from  40  to  50  silica  and  20  alumina  to  25  or  35  silica  and 
35  alumina)  with  very  little  effect  upon  the  fusibility  of  the 
slag.  On  the  other  hand,  with  a  low  proportion  of  silica  in  the 
slag,  say  30  to  40  per  cent.,  it  is  equally  possible  to  change  the 
relations  of  lime  to  alumina  within  large  limits  (i.e.,  from  50 
lime  and  15  alumina  to  35  lime  and  35  alumina)  with  very  little 
effect  upon  the  fusibility  of  the  slag. 

The  blast  furnace  manager  usually  decides  upon  the  kind 
of  slag  he  will  make,  on  one  of  three  or  four  assumptions: 

(1)  If  there  is  little  alumina  present,  and  practically  no 
magnesia,  he  usually  assumes  some  ratio  between  the  weights 
of  silica  and  lime  which  he  desires  his  slag  to  possess,  and  cal- 
culates the  weight  of  flux  necessary  to  make  slag  conforming 
to  that  condition.  In  charcoal  furnaces,  where  there  is  prac- 


254  METALLURGICAL  CALCULATIONS. 

tically  no  sulphur  in  the  charge,  the  silica  may  be  1.5  to  2.0 
times  the  lime  present;  in  coke  furnaces,  where  it  is  necessary 
to  eliminate  much  sulphur,  this  ratio  is  usually  0.5  to  1.0. 

Illustration. — If  a  ton  of  iron  ore  carries  300  pounds  of  silica, 
how  much  limestone,  which  is  practically  pure  calcium  car- 
bonate, will  be  required  to  slag  it,  neglecting  flux  necessary 
to  slag  the  ash  of  the  fuel? 

Solution. — Pure  limestone  is  CaCO3,  or  CaO  .  CO2,  and  car- 
ries 56  per  cent,  of  lime,  which  will  go  into  the  slag,  and  44 
per  cent,  of  carbonic  oxide  (CO2),  which  goes  into  the  gases. 
Using  x  pounds  of  limestone,  the  weight  of  lime  going  into  the 
slag  is  0.56  x.  If  we  assume  the  ratio  of  silica  to  lime  =  1  for 
a  coke  furnace,  and  1.75  for  a  charcoal  furnace,  we  get  the  two 
equations  and  corresponding  values  of  x: 


300 

=  1  whence  x  =  536  pounds. 


0.56* 


1.75  whence  x  =  307 


(2)  If  considerable  magnesia  is  present  it  is  usual  to  either 
count  it  simply  as  so  much  lime,  or  else  to  calculate  the  weight 
of  lime  to  which  it  is  chemically  equivalent,  and  add  this  to  the 
lime,  calling  the  sum  the  "  summated  lime."  The  ratio  of  silica 
to  lime  is  then  used  as  the  ratio  of  silica  to  lime  plus  magnesia, 
or  of  silica  to  summated  lime.  When  there  is  considerable 
magnesia  present  the  chemical  summation  should  always  be 
made.-  Small  amounts  of  MnO,  FeO  and  K2O  or  Na20  are 
also  chemically  summated  as  lime.  The  chemical  summation  is 
based  on  the  fact  that  one  molecule  of  a  base  containing  one 
atom  of  oxygen  is  considered  the  equivalent  in  fluxing  power 
of  any  other  similar  molecule;  e.g.,  CaO,  MgO,  FeO,  MnO, 
K2O,  Na20  are  considered  as  equivalent;  but  as  these  mole- 
cules weigh  differently,  we  have  equivalent  fluxing  powers  in 
the  following  weights: 

CaO 56 

MgO 40 


CALCULATION  OF  THE  CHARGE.  255 

FeO 72 

MnO... 71 

K2O 94 

Na2O 62 

from  which  we  conclude  that  since  40  parts  by  weight  of  mag- 
nesia is  equivalent  to   56  parts  of  lime,  that,  therefore,  the 

56 

lime  equivalent  of  any  weight  of  magnesia  is  —  of  the  weight 

of  the  magnesia.     Similarly,  we  get  the  lime  equivalent  of  these 
bases  as  follows: 

56 

CaO  equivalent  of  any  weight  of  MgO  =  -—X  weight  MgO 


FeO    =  ~x      «       FeO 
I  A 


MnO  =  ^X  MnO 

> 

KA 


Na20  =       X      "       Na20 
uZ 

Illustration. — An  iron  ore  contributes  to  the  slag  350  pounds 
of  silica,  12  of  FeO  and  60  of  MnO.  It  is  desired  to  flux  it  by 
using  limestone  containing  38.1  per  cent,  lime,  13.6  magnesia, 
3.4  silica,  and  44.9  carbonic  oxide  (CO2).  How  much  flux 
must  be  used  to  produce  a  ratio  of  silica  to  summated  lime 
=  0.8? 

Solution. — Letting  x  be  the  pounds  of  limestone  used,  the 
ingredients  of  the  slag  will  be 

SiO2    =  350  +  0.034  x 
CaO    =  0.381* 

MgO   =  0.136  x 

FeO    =     12 
MnO  =    60 


256  METALLURGICAL  CALCULATIONS. 

The  lime  equivalents  of  the  MgO,  FeO  and  MnO  are 

56 
CaO  equivalent  of  MgO     =  0.136  *X       =  0.1904  x 


FeO     -     12       X^  =    9.33 

MnO    =60       X^  =  47.32 
CaO     =  0.381  xX   1  =    0.381  * 


Summated  CaO  =  0.5714  #  +  56.65 

The  ratio  of  silica  to  summated  lime  is  therefore: 


350  +  0.034* 

=  U.o 


56.65  +  0.5714* 

Whence  *  =  720  pounds. 

It  is  easily  seen  that  this  method  of  solution,  calling  *  the 
weight  of  flux  used,  and  then  getting  expressions  for  the  weight 
of  each  ingredient  in  the  slag  and  the  weight  of  the  whole  slag, 
is  a  very  general  solution  which  is  applicable  to  any  kind  of 
assumed  composition  to  which  it  is  desired  that  the  slag  shall 
conform. 

(3)  If  alumina  is  present  in  the  slag-forming  constituents  of 
the  ore  it  also  may  be  reckoned  with  in  several  ways.  It  may 
be  reckoned  as  so  much  by  weight,  and  added  in  as  such  to  the 
silica  or  the  lime,  or  it  may  be  calculated  to  its  silica  or  its 
lime  equivalent,  and  added  into  the  summated  silica  or  the 
summated  lime.  Here  we  touch  on  a  question  which  has 
agitated  blast  furnace  managers  and  theorists  for  a  generation: 
Should  the  alumina  be  reckoned  with  the  bases  or  with  the 
acids;  summated  as  silica  or  as  lime?  It  would  be  presump- 
tuous to  set  forth  a  dictum  on  a  subject  which  has  been  so 
long  and  so  ably  discussed  by  some  of  the  best  iron  metal- 
lurgists, but  we  will  assume,  as  somewhere  near  the  truth,  that 
as  far  as  the  elimination  of  sulphur  in  the  slag  is  concerned, 


CALCULATION  OF  THE  CHARGE.  257 

alumina  acts  in  slags  low  in  silica  as  though  it  were  lime,  not 
in  the  proportions  of  its  lime  equivalent 


(1  AS  \ 

j02  X  weight  of  aluminaj 


but  rather  in  about  the  proportions  of  its  simple  weight.  As  far 
as  fusibility  is  concerned,  in  high  silica  slags  alumina  increases 
the  fusibility  up  to  a  certain  point,  above  which  it  decreases  it. 
It  acts  in  these,  therefore,  like  lime,  and  may  be  classed  with 
the  bases.  In  low  silica  slags-,  below  45  per  cent.,  alumina  acts 
like  silica  when  considerable  is  present,  and  like  lime  when  less 
is  present;  for  instance,  in  a  low  lime,  high  alumina  slag,  alu- 
mina and  silica  may  be  substituted  one  for  the  other  within 
wide  limits,  without  materially  affecting  the  fusibility  of  the 
slag;  in  a  high  lime,  high  alumina  slag,  alumina  and  lime  may 
be  substituted  for  each  other  within  wide  limits  without  sen- 
sibly changing  the  fusibility.  To  state  the  matter  as  succinctly 
as  possibe,  in  slags  low  in  silica  (30  to  35  per  cent.),  alumina 
reinforces  the  bases  in  the  elimination  of  sulphur;  in  regard  to 
fusibility,  it  acts  like  silica  in  a  slag  low  in  both  silica  and  lime, 
and  like  lime  in  all  other  blast  furnace  slags. 

Illustration. — An  iron  ore  carries  10  per  cent,  of  its  weight  of 
silica  and  6  per  cent,  of  alumina.  The  lime  stone  on  hand  con- 
tains 37.3  per  cent,  lime,  13.3  magnesia,  3.3  silica,  44  carbonic 
oxide  (CO2),  and  2.1  per  cent,  alumina.  How  much  flux  is 
required  per  1,000  parts  of  ore  to  make  (a)  a  slag  with  49  per 
cent,  of  silica  plus  alumina;  (b)  a  slag  with  33  per  cent,  silica; 
(c)  a  slag  with  summated  silica  =  the  summated  lime? 

Solution. 

(a)    Weight  of  SiO2  in  slag  =  100  +  0.033  x 

Weight  of  A1203  in  slag  =     60  +  0.021  x 

Weight  of  CaO  in  slag  =            0.373  x 

Weight  of  MgO  in  slag  =             0.133  x 


Total  weight  of  slag  =  160  +  0.560  x 
therefore, 

160  +  0.054*  =  0.49  (160  +  0.560  x) 
whence 

*=  370 


258  METALLURGICAL  CALCULATIONS. 

(b)  100  +  0.033  x  =  0.33  (160  +  0.560  x) 
whence 

x  =  313 

(c)  Silica  =  100        +0.033* 

180 
Silica  equivalent  of  alumina     =  —  (60  +  0.021  x) 

Summated  silica                         =153  +0.0515* 

Lime  0.373  x 

eft 

Lime  equivalent  of  magnesia  =  —  (0.133  x) 

4U 

Summated  lime  =  0.5592* 

therefore  153  +  0.0515  *  =  0.5592  * 

whence  *  =  300 

Returning  to  the  slag  resulting  from  the  proportions  of  flux 
chosen  in  Problem  51,  and  containing: 

SiO2    33.74  per  cent.  FeO    1.54  per  cent. 

APO3    5.43        "  MnO  4.21 

CaO    46.23        "  K2O    1.54       " 

MgO     7.11       "  CaS    0.20       " 

We  see  that  its  percentage  of  silica  is  low,  therefore  it  is  adapted 
to  produce  pig  iron  low  in  sulphur;  its  percentage  of  alumina 
is  low,  and  therefore  its  presence  increases  the  fusibility  of 
the  slag,  which  would  otherwise  be  rather  deficient,  because 
of  the  high  lime  and  somewhat  considerable  amount  of  other 
bases.  In  such  a  slag,  alumina  would  be  summated  with  the 
bases. 

oo    T A 

The  ratio  of  silica  to  bases  is •  _'     •  =  0.516 

ob.Ub 


The  ratio  of  silica  to  lime  plus  magnesia  is '       =  0.632 

oo .  oo 

The  summated  lime  =  46.23 

+  CaO  equivalent  of  APO3  =  ^  (5.43)  =    8.94 


CALCULATION  OF  THE  CHARGE.  259 


cc 

+  CaO  equivalent  of  MgO   =  —  (7.11)  =  9.95 

4U 


+  CaO  equivalent  of  FeO    =         (1-54)  =  1.20 


H-CaO  equivalent  of  MnO  =  ~  (4.21)  =  3.32 


+  CaO  equivalent  of  K2O    =  -^-  (1.54)  =  0.92 

70.56 

33  74 

Ratio  silica  to  summated  lime      =  ^—-^          =  0.478 

70.56 

Problem  52. 

In  a  blast  furnace  charge,  consisting  of  1530.2  pounds  of  ore 
and  682  pounds  of  charcoal,  per  1,000  pounds  of  pig  iron  made, 
it  is  known  from  the  balance  sheet  of  above  materials  that  they 
will  contribute  to  the  slag  the  following  slag-forming  ingre- 
dients. (See  balance  sheet,  Problem  51) : 

SiO2  70.9  pounds.  FeO  3.2  pounds. 

APO3  11.6        "  MnO  9.3 

CaO  39.9        "  K2O  3.4 

MgO  15.5       ."  CaS  0.4 

The  limestone  at  hand  contains : 

CaO      53.74  per  cent.  APO3       0.32  per  cent. 

MgO       0.17       "  Fe2O3       0.18       " 

SiO2        3.14       "  CO2        42.42       " 

Required. — The  weight  of  limestone  to  be  used  to  make: 
(1)  A  slag  containing  33.74  per  cent,  of  silica.  (2)  A  slag  in 
which  the  ratio  of  silica  to  bases  is  0.516.  (3)  A  slag  in  which 
the  ratio  silica  to  summated  lime  "s  0.478. 

Solution. — This  problem  embodies  the  conditions  which  con- 


260  METALLURGICAL  CALCULATIONS. 

front  the  metallurgist  when  desiring  to  calculate  the  flux  needed 
by  any  given  furnace,  and  we  have  assumed  certain  working 
ratios  to  be  aimed  at  in  the  slag,  in  order  to  elucidate  the  method 
of  solution. 


SiO2  

CONSTII 
From  On 
and  Fuel. 
70.9 

ruENTs  OF  SLAG. 
? 
From  Flux. 
0.0314* 
0.0032* 
0.5374* 
0.0017* 

|g  (0.0018*) 

Total. 
70.9  +  0.0314* 
11.6  +  0.0032* 
39.9  +  0.5374* 
15.5  +  0.0017* 

3.2  +  0.0016* 
93       

APO3 

11.6 

CaO          .... 

39.9 

MgO.. 

15.5 

FeO  

3.2 

MnO 

9  3 

K20 

3  4 

34         .    .    .    . 

CaS     .    . 

0  4 

0.4       

154.2  0.5753*        154.2  +  0.5753.* 

(1)  To  make  a  slag  with  33.74  per  cent,  of  silica  we  Must 
have 

70.9  +  0.0314*  =  0.3374  (154.2  +  0.5753*) 
whence  *  =  116  pounds. 

(2)  To  make  a  slag  with  ratio  of  silica  to  bases  0.516  we 
must  have 

70.9  +  0.0314*  =  0.516  (82.9  +  0.5439*) 
whence  *  =  116  pounds. 

(3)  To  make  a  slag  with  ratio  of  silica  to  summated  lime 
0.478,  we  must  first  summate  the  lime  as  follows: 

Lime  =  39.9  +  0.5374* 


Lime  equiv.  of  APO3  =          (11.6  +  0.0032*)      =  19.1  +  0.0053* 


MgO   =  -j-  (15.5  +  0.0017*)      =  21.7  +  0.0024* 


KC 

FeO    =  -|j(  3.2  +  0.0016*)      =     2.5  +  0.0012* 


CALCULATION  OF  THE  CHARGE.  261 

MnO   =  -|^  (  9.3)  =     7.3 

K20    =  -^  (  3.4)  =     2.0 


Summated  Ihne  =  92.5  +  0.5463  x 

therefore 

70.9  +  0.0314*  =  0.478  (92.5  +  0.5463*) 
whence  *  =  116  pounds. 

COMPARISON  OF  FUELS,  FLUXES  AND  ORES. 

By  properly  utilizing  the  preceding  principles,  it  is  possible 
to  compare  different  varieties  of  fuels,  fluxes  or  ores  with  each 
other,  and  thus  to  determine  their  relative  values  to  the  fur- 
nace, as  far  as  can  be  inferred  from  their  chemical  composi- 
tion. (Some  of  the  following  methods  are  from  a  paper  by 
Mr.  F.  W.  Gordon,  Trans.  Am.  Institute  Mining  Eng.,  1892, 
p.  61.) 

COMPARISON  OF  FUELS. 

If  different  qualities  of  fuel  are  available  it  is  possible  to 
calculate  which  is  the  most  advantageous  to  use  in  the  furnace. 
The  fixed  carbon  only  is  efficient  for  the  furnace,  and  not  all 
of  that,  because  the  ash  of  the  fuel  needs  to  be  fluxed  to  slag, 
and  a  certain  amount  of  the  fixed  carbon  will  need  to  be  burned 
simply  to  melt  this  slag.  Then  the  cost  of  the  limestone  to  be 
used  to  flux  this  ash  must  be  counted  in,  and  finally  part  of  the 
labor  costs  of  running  the  furnace  must'  be  charged  against  the 
slag.  We  can  thus  calculate  the  total  ^charges  against  the  fuel 
to  supply  one  part  of  available  carbon,  which  is  the  best  basis 
upon  which  to  compare  different  fuels. 

Illustration. — Two  varieties  of  coke  are  available  for  a  blast 
furnace,  analyzing  respectively: 

No.  1.  No.  2. 

Fixed  carbon 84  per  cent.  90  per  cent. 

Volatile  matter 2  1 

Moisture 5       "  3       " 

Ash 9       "  6       " 

And  costing  respectively  $4.50  and  $5.50  per  ton.     The  ash  of 

the  fuels  analyzes  respectively: 


262  METALLURGICAL  CALCULATIONS. 

No.  1.  No.  2. 

Silica 55  per  cent.  25  per  cent. 

Alumina 25       "  5       " 

Lime 15       "  50       " 

Magnesia 5       "  10       " 

Ferric  oxide —       "  10       " 

They  are  to  be  fluxed  with  the  limestone  of  preceding  prob- 
lem, assumed  to  cost  $1.00  per  ton,  and  to  make  a  slag  carrying 
40  per  cent,  of  silica  and  alumina  together.  Assume  an  average 
of  0.228  parts  of  fixed  carbon  necessary  to  melt  down  1  part 
of  slag,  and  that  the  manufacturing  costs  borne  by  the  slag 
amount  to  $1.00  per  ton.  What  are  the  relative  values  of  the 
two  fuels  in  this  furnace? 

Solution. — The  amounts  of  flux  needed  to  100  parts  of  each 
fuel  burned  will  be  found  as  follows,  letting  x  be  the  amount  of 
flux  used: 

Slag  No.  1.  Slag  No.  2. 

Silica 4.95  +  0.0314*         1.50  +  0.0314* 

Alumina 2.25  +  0.0032*         0.30  +  0.0032* 

Lime 1.35  +  0.5374*         3.00  +  0.5374* 

Magnesia 0.45  +  0.0017  *         0.60  +  0.0017  * 

Ferrous  oxide..  0.0016*         0.54  +  0.0016* 


Total  weights 9.00  +  0.5753  *          5.94  +  0.5753  * 

Therefore,  in  case  No.  1: 

7.20  +  0.0346*  =  0.40  (9.00  +  0.5753*) 
whence  *  =  18.4 

And  in  case  No.  2: 

1.80  +  0.0346*  =  0.40  (5.94+0.5753*) 
whence  *  =  —2.9 

The  negative  value  in  case  No.  2  simply  means  that  the  ash 
of  fuel  No.  2  is  more  basic  than  the  slag,  and,  therefore,  requires 
no  limestone,  but  itself  acts  as  a  basic  flux.  Per  ton  of  fuel 
burned,  there  would  be  required  respectively  0.184  and  —  0.029 
tons  of  limestone,  and  the  weights  of  slag  would  be  0.196  tons 
and  0.0427  tons.  (Substituting  values  of  *  in  the  total  weights 
of  slags.) 

The  weights  of  fixed  carbon  necessary  to  smelt  these  weights 
of  slag  would  be: 


CALCULATION  OF  THE  CHARGE.  263 

No.  1.     0.196   X0.228  =  0.0447  tons. 
No.  2.     0.0427X0.228  =  0.0097     " 

Leaving  as  available  fixed  carbon  for  the  furnace  in  each  case: 

No.  1.     0.84-  0.0447  =  0.7953  tons. 
No.  2.     0.90-  0.0097  =  0.8903     " 

From  these  figures  the  cost  of  1  ton  of  available  carbon  fur- 
nished by  each  fuel,  adding  in  manufacturing  cost  chargeable 
against  the  slag,  is 

No.  1. 

Cost  of  coke,  $4.50  ^  0.7953  =  $5 . 658 

Cost  of  limestone,  $1.00X0.184^0.7953  =    0.231 

Costs  against  slag,  $1.00X0. 196 -r- 0.7953  =    0.246 

$6.135 
No.  2. 

^Cost  of  coke,  $5.50-^0.8985  =  $6.233 

Cost  of  limestone,  $1.00 X (-  0.092)  +  0.8903    =  -0. 101 
Cost  against  slag,  $1.00X0.0065X0.8903         =    0.007 


$6.139 

The  two  fuels,  at  the  prices  given,  are  therefore  of  almost 
exactly  the  same  value  to  the  furnace. 

The  solution  along  the  lines  shown  is  general,  for  any  de- 
sired composition  of  slag,  or  for  use  with  any  given  limestone, 
and  is  a  valuable  means  of  comparing  the  values  of  different 
fuels.  The  cost  of  a  ton  of  puie,  available  carbon,  when  fur- 
nished by  any  given  fuel,  is  an  item  which  is  useful  when  com- 
paring the  relative  values  of  different  fluxes  or  ores  with  each 
other. 

COMPARISON  OF  FLUXES. 

If  different  qualities  of  flux  are  available  it  is  very  desirable 
to  be  able  to  calculate  which  is  the  most  economical  to  use  in 
the  furnace.  Any  acid  ingredients  in  the  flux  diminish  very 
sharply  its  efficient  fluxing  power,  because  they  must  first  be 
satisfied  from  the  bases  present  in  the  same  proportions  as  acid 
to  bases  in  the  final  slag.  The  slag  thus  formed  from  the  im- 
purities requires  further  to  be  melted,  and  other  costs  are 
properly  chargeable  against  it.  The  best  comparison  is  finally 


264  METALLURGICAL  CALCULATIONS. 

obtained  by  calculating  for  each  flux  available  the  cost  from  it 
of  pure  net  lime,  or  net  summated  lime,  analogous  to  the  cal- 
culation for  pure  net  carbon  in  the  case  of  fuels.  Any  ordinary 
condition  may  be  imposed  upon  the  slag  which  the  furnace  is 
to  produce. 

Illustration. — There  are  available  for  a  furnace  two  qualities 
of  limestone,  containing  respectively: 

CaO 53.74  per  cent.  47.80  per  cent. 

MgO 0.17       "  4.61       " 

SiO2 3.14       "  5.12       " 

APO3 0.32       "'  3.36 

Fe203 0.18       "  1.10       " 

CO2 42.42       "  37.55       " 

The  first  costs  $1.00  per  ton,  the  second  $0.80.  Assume  them 
smelted  with  fuel  furnishing  pure  available  fixed  carbon  at 
$6.135  per  ton ;  that  0.228  tons  of  pure  fixed  carbon  is  needed  to 
smelt  1  ton  of  slag;  that  manufacturing  costs  against  slag  are 
$1.00  per  ton,  and  that  the  slag  to  be  made  in  the  furnace  must 
have  summated  silica  equal  to  summated  lime.  Compare  the 
relative  values  of  the  two  fluxes. 

Solution. — We  will  direct  our  calculations  towards  finding 
the  net  cost  of  1  ton  of  pure  available  summated  lime  from 
each  of  the  two  limestones.  The  summated  lime  and  silica  in 
each  flux  are: 

No.  1.  No.  2. 

Summated  lime. 0.5411  0.5502 

Summated  silica . .  .  .  0 . 0342  0 . 0808 


Excess  of  summated  lime 0 . 5069  0 . 4694 

The  weights  of  slag  formed  from  the  impurities  present  in 
each  limestone  will  be: 

No.  1.  'No.  2. 
CaO  (difference  between  amount  present 

and  excess  of  summated  lime  found) .  .  0 . 0305  0 . 0086 

MgO 0.0017  0.0461 

FeO 0.0016  0.0099 

SiO2 0.0314  0.0512 

APO3..                                                        ..0.0032  0.0336 


Totals..  ..0.0684  0.1494 


CALCULATION  OF  THE  CHARGE  265 

The  cost  of  1  ton  of  pure  available  lime  from  each  of  thes<5 
fluxes  will  therefore  be,  adding  in  costs  chargeable  against  the 
slags  formed  by  the  impurities  present: 

No.  1. 

Cost  of  limestone,  $1.00^0.5069  =  $1.973 

Cost  of  carbon  for  melting  slag,  $6.135X0.228X0.0684 

-^  0.5069  =    0.188 

Costs  of  running,  chargeable  against  slag,  $1.00  X  0.0684 

-T- 0.5069  =0.135 


No.  2. 

Cost  of  limestone,  $0.80-^0.4694 
Cost  of  carbon  for  melting  slag,  $6.135X0.228X0.1494 

^0.4694  =    0.465 

Cost  of  running,  chargeable  against  slag,  $1.00X0.1494 

4-0.4694  =    0.318 


$2.487 

The  conclusion  is  that  the  poorer  limestone,  at  $0.20  per  ton 
less  cost,  is  in  reality  costing  $0.191  per  ton  more  for  pure 
available  lime,  or  is  in  reality  8.3  per  cent,  dearer  than  the  first, 
instead  of  being  20  per  cent,  cheaper. 

The  method  of  calculation  here  described  is  quite  general  for 
any  compositions  of  limestone  or  other  flux,  and  for  any  as- 
sumed conditions  which  the  slag  must  conform  to. 

COMPARISON  OF  ORES. 

As  the  more  complicated,  we  come  to  the  comparison  of 
various  ores  which  may  be  at  the  iron  master's  disposal.  Here 
a  similar  method  of  procedure  is  advisable.  It  can  be  calcu- 
lated first,  for  a  unit  weight  of  ore,  how  much  pure  lime  would 
be  required  to  flux  its  impurities,  how  much  pure  carbon  would 
be  required  to  melt  the  slag  thus  formed,  and  to  the  costs  of 
each  of  these  would  be  added  the  handling  of  the  slag.  Each 
of  these  can  be  also  expressed  per  unit  of  pure  oxide  of  iron 
in  the  ore ;  and  if  to  their  sum  we  add  the  cost  of  ore  necessary 
to  furnish  unit  weight  of  pure  oxide  of  iron  we  obtain  the  total 
costs  per  unit  weight  of  pure  iron  oxide  (Fe2O3).  This  is  the 
basis  on  which  different  ores  may  then  be  compared.  It  must 
not  be  forgotten  that  one  ore  may,  because  of  higher  sulphur 


266  METALLURGICAL  CALCULATIONS. 

content,  require  the  production  of  a  more  basic  slag,  so  that  the 
amounts  of  flux  required  and  slag  formed  will  be  influenced 
by  the  condition  necessary  to  impose  on  the  slag  in  each  case. 

Illustration.  —  The  iron  ore  briquettes  (of  Problem  51)  con- 
tained Fe2O3,  85.93  per  cent.;  FeO,  3.96;  SiO2,  5.50;  MnO,  0.63; 
A12O3,  0.76;  CaO,  2.23;  MgO,  0.97  per  cent.  If  these  cost  $4.40 
per  ton,  and  are  smelted  in  a  furnace  making  slag  with  ratio 
of  silica  to  bases  0.516,  and  assuming  0.3  per  cent,  of  the  iron, 
82.7  per  cent,  of  the  silica,  and  96.6  per  cent,  of  the  manga- 
nese to  go  into  the  slag,  what  is  the  cost  per  ton  of  pure  Fe2O3 
from  this  source,  charging  pure  lime  for  fluxing  at  $2.296  per 
ton,  pure  carbon  for  smelting  slag  at  $6.135  per  ton,  and  re- 
quiring 0.228  tons  of  carbon  for  one  of  slag,  adding  also  manu- 
facturing costs  at  $1.00  per  ton  of  slag? 

Solution.  —  The  slag-forming  ingredients  from  1  ton  of  ore 
briquettes  are: 


FeO  0.003  x       X 
55 


|7 


o.8593X  ~      +    o.0396x  =  0.0024  tons 


MnO  0.966X0.0063  =0.0061 

CaO  =  0.0223 

MgO  =  0.0097 

A12O3  =  0.0076 

SiO2  =  0.0455 

and  the  bases  to  satisfy  the  silica  present  must 

be  0.0455  -v-  0.516  ,             =0.0882 

But,  sum  of  bases  already  present  =  0.0482 

therefore,  pure  lime  to  be  added  =  0.0417 

and  total  weight  of  slag  =  0.1337 

Efficient  Fe2O3  in  1  ton  of  ore  =  0.8593 

plus  Fe2O3  equivalent  of  reduced  FeO 

Ifin 
=  ~     (0.0396-  0.0024)=  0.0413 


Total  available  Fe2O3  =  0.9006     u 


METALLURGICAL  CALCULATIONS.  267 

Cost  of  1  ton  pure  available  Fe2O3  from  these  briquettes: 
Cost  of  ore,  $4.40  -*-  0.9006  =  $4 . 885 

Cost  of  pure  lime,  $2.296x0.0417-^0.9025  =    0.106 

Cost  of  carbon  for  melting  slag,  $6.135X0.228X0.1337 

-0.9025  =    0.207 

Costs  chargeable  against  slag,  $1.00X0.1337-=- 0.9025  =    0.148 

Total  =  $5.346 

A  similar  calculation  is  possible  with  any  ore  of  any  given 
composition,  and  making  any  assumed  quality  of  slag.  The 
costs  of  1  ton  of  pure  ferric  oxide  thus  calculated  will  give  the 
relative  costs  of  the  iron  obtained  from  these  different  sources, 
and  therefore  indicate  the  relative  values  of  the  different  ores 
to  the  blast  furnace  manager. 

Problem  53. 

Assume  a  blast  furnace  manager  to  use  the  ore  of  preceding 
illustration,  furnishing  pure  Fe2O3  at  a  net  cost  of  $5.336  per 
ton,  and  to  use  with  it  fuel  furnishing  pure  carbon  at  $6. 135  per 
ton,  there  being  required  for  reduction  and  melting  the  iron 
produced  and  furnishing  it  with  carbon,  0.66  tons  of  pure 
available  carbon  per  ton  of  pig  iron  produced,  and  the  pig 
iron  containing  96.656  per  cent,  of  iron.  The  running  costs  of 
the  furnace  are  $3.00  per  ton  of  pig  iron  produced  (costs  against 
slag  not  included) . 

Required: — The  cost  of  the  pig  iron  per  ton. 

Solution: 

Fe203  required  160  -5- 112  X  0.96656  =  1 . 3808  tons. 

Cost  of  the  ore,  $5.336 X  1.3808  =  $7.368 

Cost  of  fuel,  $6.135X0.66  =    4.049 

Cost  of  manufacturing  (share  against  pig  iron)  =    2 . 000 

Total  cost  =  $13.417 


CHAPTER  III. 
UTILIZATION  OF  FUEL  IN  THE  BLAST  FURNACE. 

The  blast  furnace,  in  its  simplest  terms,  may  be  regarded  as 
a  huge  gas  producer,  producing  in  the  region  of  the  tuyeres 
pure  producer  gas  from  fixed  carbon  and  heated  air;  the  gas 
thus  produced  is  partly  oxidized  in  its  ascent  through  the 
furnace  by  the  oxygen  abstracted  from  the  charge  (which 
latter  item  is  almost  a  constant  quantity  per  unit  of  pig  iron 
made),  and  has  added  to  it  carbon  dioxide  from  the  carbonates 
of  the  charge.  But,  after  all,  the  unoxidized  and  combustible 
ingredients  of  the  gas  escaping  represent  a  large  part,  in  fact, 
often  the  largest  part,  of  the  total  calorific  power  of  the  fuel. 

Problem  54. 

A  blast  furnace  uses  2,240  pounds  of  coke,  containing  90  per 
cent,  fixed  carbon  and  350  pounds  of  limestone,  containing  10 
per  cent,  of  carbon  (as  carbonic  acid,  CO2)  to  produce  a  ton  of 
pig  iron  containing  4  per  cent,  of  carbon.  The  gases  contain 
24  per  cent,  of  carbonous  oxide,  CO,  12  per  cent,  of  carbonic 
oxide,  CO2,  2  per  cent,  of  hydrogen,  2  per  cent,  of  methane,  and 
60  per  cent,  of  nitrogen. 

Required. — (1)  The  volume  of  gas,  as  analyzed,  produced  per 
ton  of  pig  iron  made. 

(2)  The  calorific  power  of  the  gas. 

(3)  The  proportion  of  the  calorific  power  of  the  coke  which 
has  been  generated  in  the  furnace. 

Solution. — (1)  The  carbon  going  into  the  gases  will  be  that 
in  the  coke,  less  that  in  the  pig  iron,  plus  that  in  the  carbonates 
of  the  charge. 

Carbon  in  coke  =  2,240X0.90  =  2,016     pounds 

Carbon  in  carbonates  =      350X0.10=        35 
Carbon  charged  =  2,051 

Carbon  in  pig  iron  =  2,240X0.04  =        89.6      " 

Carbon  going  into  the  gases  =  1,961.4      " 

268 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          269 

Carbon  in  1  cubic  foot  of  gas: 
In  CO    0.24X0.54 
In  CO2   0.12X0.54 
In  CH4  0.02X0.54 


Total  0.38X0.54  =  0.2052  ounces  av. 
=  0.012825  pounds 

Gas  produced  per  ton  of  pig  iron  =     '       '      =  152,935  cu.  ft.(l) 


(2)  Calorific  power  of  1  cubic  foot  of  gas: 
CO    0.24X3,062  =  734.9  oz.  cal. 
H2     0.02X2,613  =    52.3       " 
CH4  0.02X8,598  =  172.0       " 


Sum  =  959.2 

=    59.95  pound  cal. 
per  152,935  cubic  feet  =  9,168,450  pound  cal.  (2) 

(3)  The  calorific  power  of  the  coke  considering  it  to  contain 
simply  90  per  cent,  of  fixed  carbon,  would  be 

8,100X0.90  =  7,290  pound  cal.  per  pound. 

The  presence  of  CH4  in  the  gases  points,  however,  to  there 
being  probably  some  available  hydrogen  in  it,  which  would  in- 
crease its  calorific  power  somewhat.  A  closer  approximation 
to  the  calorific  power  of  the  coke  could,  therefore,  be  obtained 
by  assuming  at  least  as  much  available  hydrogen  in  it  as  would 
correspond  to  the  hydrogen  in  the  CH4  in  the  gas. 

CH4  in  152,935  cu.  ft.  of  gas  3,059  cu.  ft. 

Weight  of  this  =  3,059  X  (0.09  X  8)      =  2,202  oz.  av. 

137.7  Ibs. 

Hydrogen  =  137.7X  (4-J-16)  34.4  Ibs. 
Available  hydrogen  in  coke  =  34.4 

-s-  2,240  1.54  percent. 

Calorific  power  0.0154X29,030  447  Ib.  cal.  per  Ib. 

Total  calorific  power  of  the  coke  7,737     "         " 
Calorific  power  of  coke  used  per  ton 

7,737X2,240                                     =  17,330,880  Ib.  cal. 

Calorific  power  of  the  gases                 =  9,168,450     " 

Calorific  power  generated  =    8,162,430     " 

=  47.1  per  cent. 


270  METALLURGICAL  CALCULATIONS. 

This  figure,  however,  practically  over-charges  the  furnace 
little  bit,  because  the  pig  iron  with  its  4  per  cent,  of  carbon 
really  takes  out  of  the  furnace  some  unburnt  fuel,  whose  heat 
of  combustion  may  be  utilized  outside  the  furnace  as  in  the 
Bessemer  converter.  The  furnace  does  not  generate  heat  from 
this,  representing: 

2,240X0.04X8,100  =  725,760  Ib.  cal. 
leaving  as  actually  generated  in  the  furnace 

8,162,430-  725,760  =  7,436,670  Ib.  cal. 

=  42.9  per  cent.  (3) 

Such  an  average  blast  furnace  cannot,  therefore,  be  accused 
of  generating  within  it  over  some  43  per  cent,  of  the  calorific 
power  of  the  fuel  put  into  it,  while  the  heat  rejected  as  poten- 
tial energy  of  combustion  of  the  waste  gases  amounts  to  more 
than  half  the  calorific  power  of  the  fuel.  Fifty  years  or  more 
ago,  when  these  waste  gases  were  allowed  to  burn  truly  to  waste 
the  blast  furnace  was  indeed  a  devourer  of  fuel,  but  matters 
have  been  improved  by  the  utilization  of  the  waste  gases  to  heat 
the  blast,  and  thus  one  of  the  largest  "leaks"  of  heat  from 
the  furnace  has  been  patched  up  to  some  extent,  although  yet 
far  from  satisfactorily. 

Problem  55. 

Assume  that  in  problem  54,  one-third  of  the  gases  produced 
are  burnt  in  hot-blast  stoves,  preheating  the  air  blown  in  at  the 
tuyeres,  and  that  the  blast  is  thus  preheated  to  450°  C. 

Required. — (1)  The  amount  of  blast  blown  in  per  ton  of  pig 
iron  made. 

(2)  The  heat  in  the  blast. 

(3)  The  efficiency  of  the  hot-blast  stoves. 

(4)  The  increased  efficiency  of  the  blast  furnace  plant  as  a 
whole  in  generating  the  calorific  power  of  the  fuel,  when  thus 
provided  with  this  hot-blast  apparatus. 

Solution. 

(1)  Volume  of  (dry)  gases  per  ton      =  152,935  cu.  ft. 
Nitrogen  present  in  these  (60%)  =    91,761 

Air  containing  this  =  ^~        =  115,860       " 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          271 

This  is  the  volume  of  the  blast  per  ton  of  pig  iron  produced, 
assuming  no  nitrogen  to  come  from  the  coke  used,  and  the 
blast  to  be  dry.  If  the  blast  were  moist,  and  its  hydrometric 
condition  known,  the  volume  of  moist  blast  could  be  calculated. 

(2)  Assuming  the  blast  dry,  it  is  heated  to  450°  C.,  requiring 

1  15,860  X  [0.303  +  0.000027  (450)]X450    =  16,430,975  oz.  cal. 

=     1,026,936  Ib.  cal. 

(3)  The  hot-blast  stoves  receive  one-third  of  all  the  gas  pro- 
duced, having,  therefore,  a  calorific  power  of 

9,168,450-^3  =  3,056,150  Ib.  cal. 


Efficiency  of  the  stoves  =  o''n  =  °-336  =  33-6  Per  cent- 

6,  (Job,  150  /3\ 

(4)  The  blast  furnace  was  primarily  rejecting  unused  57.1 
per  cent,  of  the  calorific  power  of  the  fuel,  4.2  per  cent.,  how- 
ever, as  a  necessary  loss,  to  supply  the  carbon  in  the  pig  iron, 
but  52.9  per  cent,  as  combustible  power  of  unburnt  waste  gases. 
If  one-third  of  these  gases  are  completely  burnt  in  hot-blast 
stoves,  then  the  combined  plant  —  furnace  plus  stoves  —  is 
utilizing  17.6  per  cent,  more  of  the  calorific  power  of  the  fuel 
than  before,  or  42.9  +  17.6  =  60.5  per  cent.,  and,  therefore,  re- 
jecting undeveloped  39.5  per  cent,  of  the  calorific  power.  (4) 

[The  net  effect  of  the  use  of  the  hot-blast  stove,  upon  the 
heat  generation  in  the  furnace,  is  practically  to  put  back  into 
the  furnace,  as  sensible  heat,  1-3X33.6  =  11.2  per  cent,  of  the 
calorific  power  of  the  waste  gases,  equal,  therefore,  to  0.112X 
52.9  =  5.9  per  cent,  of  the  total  calorific  power  of  the  fuel. 
This  renders  available,  for  the  working  of  the  furnace,  42.9  + 
5.9  =  48.8  per  cent,  of  the  calorific  power  of  the  fuel,  an  in- 

5  9 

crease  of  available  heat  for  reducing  and  smelting  of  777-7:  = 

4  L  .  y 

13.7  per  cent,  of  the  former  available  quantity.] 

The  practical  conclusion  is  that  a  blast  fuinace  generates 
in  itself  not  much  over  40  per  cent,  of  the  calorific  power  of 
the  fuel  used,  and  rejects  nearly  60  per  cent.  ;  by  using  part  of 
the  waste  gases  to  heat  the  blast,  however,  some  of  this  re- 
jected heat,  to  an  amount  representing  net  5  to  10  per  cent,  of 
the  calorific  power  of  the  fuel  used,  is  returned  to  and  injected 


272  METALLURGICAL  CALCULATIONS. 

bodily  into  the  furnace,  thus  rendering  available  for  the  pur- 
poses of  running  the  furnace  some  50  per  cent,  of  the  calorific 
power  of  the  fuel  as  a  maximum.  The  efficiency  with  which 
the  furnace  applies  this  50  per  cent,  usefully  to  the  objects  of 
reducing  and  smelting,  is  another  question  for  investigation. 

Problem  56. 

Assume  that  at  the  furnace  of  Problems  54  and  55  the  two- 
thirds  of  the  waste  gases  are  burnt  under  boilers,  raising  steam 
which  *runs  the  blowing  engines,  hoists  and  pumps,  and  pro- 
viding 10  effective  horse-power  for  each  ton  of  pig  iron  made 
per  day  in  the  furnace. 

Required. — (1)  The  efficiency  of  development  of  the  calorific 
power  of  the  fuel  in  the  plant  (furnace,  stoves,  boilers,  engines) 
regarded  as  a  whole. 

(2)  The  thermo-mechanical  efficiency  of  the  boiler  and  engine 
plant. 

(3)  The  power  which  could  be  generated  if  gas  engines,  at 
25  per  cent,  thermo-mechanical  efficiency,  were  used  in  their 
stead. 

Solution. — (1)  Since  the  stoves  completely  burn  one-third 
of  the  waste  gases,  and  the  boilers  the  other  two-thirds,  all  the 
combustible  power  of  the  waste  gases  is  developed  in  the  com- 
bined plant,  and  the  only  part  of  the  calorific  power  of  the 
fuel  which  is  unused  is  the  4.9  per  cent,  represented  by  the 
carbon  necessarily  entering  into  the  composition  of  the  pig 
iron.  The  plant  as  a  whole,  therefore,  develops  or  generates 
95.1  per  cent,  of  the  calorific  power  of  the  fuel  used. 

(2)  For  each  ton  of  pig  iron,  the  heat  developed  under  the 
boilers  will  be  two-thirds  of  the  calorific  power  of  the  gases,  or 

9,168,450X2-3  =  6,112,300  Ib.  cal. 

There  is  generated  thereby  10  effective  horse-power  days, 
equal  to 

10X33,000X60X24  =  475,200,000  ft.  Ibs 
But  1  Ib.  cal.  =  425X3.2808  =  1394.3      " 

Therefore,  thermal  equivalent  of 

,    -  475,200,000 

work  done  =  —  '       '  -*         340,800  Ib.  cai. 

. o 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          273 

Thermo-mechanical  efficiency  of  boiler  and  engine  plant: 
340,800 


6,112,300 

(3)  Gas  engines,  at  25  per  cent,  thermo-mechanical  efficiency, 
would  give  power  representing  per  ton  of  pig  iron  produced : 

6,112,300X       0.25  =  1,528  075  Ib.  cal. 
Equal  to      1,528,075X1394.3    =  2,130,645,900  ft.  Ibs. 

2,130,645,900  ,  -  ^ 

33,000X60X24  " 

A  quicker  solution  is: 

10  X^  =  44.9        "         "  (3) 

Leaving  net  surplus  power  per  ton  of  pig  iron  produced  per 
day  =  34.9  horse-power. 

The  preceding  problems  have  elucidated  the  question  of  the 
small  proportion  of  the  calorific  power  of  the  fuel  which  is 
generated  in  a  blast  furnace,  showing  it  to  be,  in  usual  practice 
only  40  to,  at  most,  50  per  cent,  of  the  calorific  power.  The 
discussion  has  not  explained  "  why,"  but  a  further  consideration 
will  throw  light  on  this  question  also. 

The  proportion  of  the  calorific  power  of  a  fuel  which  is  gen- 
erated in  a  blast  furnace  is  solely  a  question  of  how  much  of 
it  is  burned  to  carbonic  oxide,  CO2,  and  how  much  to  carbonous 
oxide,  CO?  If  all  the  carbon  were  burned  to  CO2,  practically 
all  the  calorific  power  of  the  fuel  would  be  generated;  if  all 

2  430 

were  burned  to  CO,  only  ~r^=  0.30  =  30    per    cent,    of    the 

o,IUU 

heating  power  of  the  carbon  would  be  generated.  If  a  blast 
furnace  was  filled  with  nothing  but  coke,  and  air  blown  in  as 
usual  at  the  tuyeres,  carbon  would  be  burned  in  the  furnace 
only  to  CO,  and  but  30  per  cent,  of  its  calorific  power  be  gen- 
erated and  available  for  the  needs  of  the  furnace.  The  entire 
gain  over  this  percentage  is  due  to  the  oxidation  of  CO  to  CO2 
by  the  oxygen  abstracted  from  the  solid  charges,  that  is,  by 
the  act  of  reduction.  In  Problem  54  we  calculated  that  under 
ordinary  conditions,  between  40  and  50  per  cent,  of  the  calorific 


274  METALLURGICAL  CALCULATIONS. 

power  of  the  fuel  is  generated  in  the  furnace;  the  excess  of 
this  above  30  per  cent,  is  due  to  the  oxidation  of  CO  to  CO2 
during  the  reduction  of  the  metallic  oxides  in  the  charge.  From 
this  standpoint  it  is  advisable  to  strive  to  perform  the  greatest 
possible  proportion  of  the  reduction  in  the  furnace  by  CO  gas, 
because  in  this  case  the  total  generation  of  heat  in  the  furnace 
per  unit  of  fuel  charged  will  tend  towards  a  maximum.  Since 
no  carbon  can  be  burned  to  CO2  at  the  tuyeres,  it  follows  that, 
from  the  standpoint  of  the  generation  of  the  maximum  quantity 
of  heat  in  the  furnace,  from  a  given  weight  of  fuel,  Gruner  was 
right  in  formulating  his  dictum  of  the  ideal  working,  of  a  blast 
furnace,  viz.: 

GRUNER'S  "  IDEAL  WORKING." 

All  the  carbon  burnt  in  the  furnace  should  be  first  oxidized 
at  the  tuyeres  to  CO,  and  all  reduction  of  oxides  above  the 
tuyeres  should  be  caused  by  CO,  which  thus  becomes  CO2. 
This  dictum  is  not  in  Gr  liner's  own  words,  but  expresses  their 
sense,  and  from  the  point  of  view  of  the  present  discussion,  it 
is  the  correct  principle  upon  which  to  obtain  the  maximum  gen- 
eration of  heat  in  the  furnace  from  a  given  weight  of  fuel.  It 
practically  directs  us  to  generate  at  the  tuyeres  30  per  cent, 
of  the  calorific  power  of  the  carbon  oxidized  in  the  furnace, 
and  the  rest  that  can  be  obtained  from  the  carbon  is  to  be 
generated  during  the  reduction  of  the  charge. 

If  we  apply  this  principle  to  the  furnace  and  data  of  Problem 
54,  we  should  first  observe  that  the  carbon  oxidized  in  the 
furnace  is: 

Carbon  in  coke  charged 2016 .     pounds 

Carbon  in  pig  iron  produced 89 . 6     .  " 

Carbon  oxidized  in  furnace 1926. 6 

Requiring,  if  all  oxidized  to  CO  at  the  tuyeres 

4 
1926. 6  X-o-  Ibs.  oxygen  =  2569.  pounds. 

u 

But,  Problem  52  shows  us  that  there  was  actually  blown 
into  this  furnace  115,860  cubic  feet  of  blast,  containing,  there- 
fore, 

115,860X0.208X(1.44-*-16)  =  2,169  Ibs.  oxygen, 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          275 

capable  of  oxidizing  to  CO  at  the  tuyeres 

2,169X0.75  =  1,627  Ibs.  carbon. 
Proportion  of  carbon  gasified  burnt  at  the  tuyeres, 
1,627 


1926.6 


0.844  =  84.4  per  cent. 


It  is,  therefore,  true  of  the  furnace  under  discussion,  that  if 
Gruner's  ideal  working  be  called  standard,  this  furnace  attains 
to  84.4  per  cent,  of  that  ideal;  and  it  is  also  true  that  this  fur- 
nace generates  from  the  carbon  burnt  at  the  tuyeres  84.4  per 
cent,  of  the  amount  of  heat  which  could  have  been  generated  if 
Gruner's  ideal  working  had  been  attained. 

It  is  always  possible  to  find  out  for  any  given  blast  furnace, 
by  similar  calculations,  how  much  carbon  is  burned  at  the 
tuyeres,  and  how  much  is  burned  above  the  tuyeres,  and  thus 
to  determine  how  closely  the  furnace  running  approximates  to 
Gruner's  ideal  working.  This  proportion  or  percentage  will 
not  necessarily  express  how  efficiently  the  furnace  is  running, 
as  regards  fuel  used  per  unit  of  iron  made,  but  it  will  tell  what 
proportion  of  the  calorific  power  of  the  fuel  used  is  being  gen- 
erated at  the  tuyeres,  and  in  possibly  nine  cases  out  of  ten 
this  proportion  indicates  the  general  efficiency  of  the  furnace 
as  regards  fuel  consumption. 

It  will  be  next  profitable  to  inquire  when  and  under  what 
conditions  Gruner's  ideal  working  does  not  correspond  to 
maximum  fuel  economy,  and  why  it  usually  does.  The  answer 
is  not  difficult  to  understand:  if  all  the  carbon  gasified  in  the 
furnace  is  burned  to  CO  at  the  tuyeres,  30  percent  of  the  total 
calorific  power  of  the  carbon  burned  is  there  developed,  which 
is  more  than  half  of  all  the  heat  generated  from  carbon  in  the 
furnace.  To  this  must  also  be  added  the  sensible  heat  in  hot 
blast,  which  may  amount  (as  in  Problem  52)  to  some  5.9  per 
cent,  of  the  calorific  power  of  the  carbon,  making,  therefore,  a 
total  of  35  per  cent,  of  the  calorific  energy  of  the  fuel  gen- 
erated at  the  tuyeres  out  of  a  total  of  about  50  per  cent,  de- 
veloped in  the  furnace.  If,  however,  the  blast  be  heated  to  a 
very  high  temperature,  or  particularly  if  it  be  dried,  or  if  the 
ore  and  fuel  are  extra  pure,  so  that  a  smaller  quantity  of  heat 
is  needed  to  melt  down  slag  at  the  tuyeres,  then  there  may  not 


276  METALLURGICAL  CALCULATIONS. 

be  needed  at  the  tuyeres  the  generation  by  combustion  of  so 
much  heat  as  Griiner's  ideal  working  would  require  and  cause 
to  be  produced,  and  to  burn  at  the  tuyeres  all  the  carbon  oxi- 
dized in  the  furnace  would  be  wasteful  of  fuel.  In  this, case 
although  less  heat  would  be  generated  per  unit  of  fuel,  by 
burning  some  of  it  above  the  tuyeres,  yet  economy  in  fuel  con- 
sumption as  a  whole  would  be  attained,  because  of  the  better 
distribution  of  the  heat  which  was  generated  from  a  smaller 
total  quantity  of  fuel. 

Illustration. — In  Problem  55  we  assumed  that  the  furnace 
ran  with  blast  heated  to  450°  C.,  and  that  this  hot  blast,  burn- 
ing at  the  tuyeres  84.4  per  cent,  of  all  the  carbon  gasified  in  the 
furnace,  smelted  down  pig  iron  and  slag  satisfactorily  and  kept 
the.  tuyere  region  at  proper  temperature.  If  the  temperature 
of  this  blast  were  raised  to  900°  C.,  how  much  greater  pro- 
portion of  heat  would  be  available  in  the  tuyere  region? 

We  have  already  calculated  that  the  115,860  cubic  feet  of  blast 
used  per  ton  of  iron  made,  brought  in  at  450°  C.,  1,026,936 
pound  calories  of  heat,  equal  to  5.9  per  cent,  of  the  calorific 
power  of  the  fuel  put  into  the  furnace.  If  the  temperature 
were  900°  C.,  the  heat  brought  in  would  be 

115,860 X [0.303 +  0.000027  (900)] X 900         =  37,920,978  oz.  cal. 

=    2,370,061  Ib.  cal. 
which  equals 

=  0.137  -  13.7  per  cent. 

of  the  calorific  power  of  the  fuel,  a  gain  of  7.8  per  cent,  added 
to  the  heat  available  in  the  tuyere  region.  This  causes  a  very 
great  increase  in  the  smelting-down  power  of  the  furnace, 
enabling  the  same  work  per  ton  of  ore  smelted  to  be  done  with 
much  less  consumption  of  fuel  in  the  tuyere  region.  An  idea 
of  this  increased  smelting-power  may  be  obtained  from  the 
following  comparison  of  heat  available  for  smelting  purposes 
in  the  tuyere  region  in  the  two  cases  just  discussed: 

CASE  1. 
Heat  developed  by  oxidation  of  carbon  per 

ton  of  iron  made,  1,627  lbs.X2,430  =  3,953,610  Ib.  cal. 

Sensible  heat  in  blast  at  450°  C.  =  1,026,936 


Total  heat  available  =  4,980,546 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          277 

CASE  2. 
Heat   developed   by   same   quantity   of   air 

burning  same  carbon  =  3,953,610       " 

Sensible  heat  in  blast  at  900°  C.  =  2,370,061 


Total  heat  available  =  6,323,671 

It  is,  therefore,  seen  that  the  heat  generated  and  available 
at  the  tuyeres  is  increased  1,343,125  calories,  amounting  to  27 
per  cent,  of  the  amount  disposable  in  Case  1.  It  follows,  there- 
fore, that  the  smelting  down  power  has  been  increased  27 
per  cent.,  and  that,  if  the  4,980,546  calories  were  sufficient  for 
satisfactorily  smelting  down  the  iron  and  slag  in  the  first  in- 
stance, that  the  extra  heat  of  Case  2  can  all  be  utilized  for 
smelting  down  27  per  cent,  extra  burden.  We  can,  there- 
fore, charge  27  per  cent,  more  burden  per  unit  weight  of  coke 
in  Case  2,  because  we  have  the  requisite  smelting  down  power 
at  the  region  of  the  tuyeres,  which  amounts  to  saying  that 
we  can  charge  22  per  cent,  less  coke  for  a  given  weight  of  pig 
iron  made. 

In  actual  practice,  as  the  amount  of  burden  is  increased  and 
the  temperature  of  the  blast  increased,  the  change  causes  more 
and  more  of  the  carbon  to  be  oxidized  above  the  tuyeres,  and 
a  smaller  proportion  to  be  oxidized  at  the  tuyeres,  thus  ob- 
taining less  service  in  the  furnace  from  oxidation  of  carbon  as 
a  whole,  but  compensating  for  this  by  the  extra  heat  in  the 
hot  blast.  Or,  looking  at  it  in  another  way,  we  may  say  that 
the  same  heat  could  be  made  available  in  the  region  of  the 
tuyeres,  when  using  hot  blast,  by  the  combustion  there  of  a 
smaller  quantity  of  carbon;  therefore,  we  can  burn  more  of  it 
above  the  tuyeres  and  yet  work  more  economically  on  the 
whole,  than  we  were  working  in  the  first  instance,  with  the 
colder  blast. 

MINIMUM  CARBON  NECESSARY  IN  THE  FURNACE. 
Many  writers  have  assumed  that  in  the  reduction  of  iron, 
oxide,  such  as  Fe2O3,  the  reaction  of  its  reduction  by  carbonous 
oxide,  CO,  is  expressed  as  follows: 

Fe2O3  +  SCO  =  2Fe  +  3CO2 
If  this  were  true,  there  would  need  to  be  burnt  to  CO  at  the 


278  METALLURGICAL  CALCULATIONS. 

tuyeres  only  3C,  or  36  parts  of  carbon,  to  ensure  the  reduction 
of  Fe2O3,  representing  112  parts  of  iron.  The  reaction  does 
not,  however,  progress  as  shown,  because  CO2  acts  oxidizingly 
on  Fe  to  such  a  degree  that  when  ICO2  is  present  in  the  gas 
for  ICO  left  unused,  the  reduction  practically  stops,  even 
though  the  gases  are  moving  slowly  through  the  warm  ore. 
The  real  reaction  of  reduction  by  CO  gas  is  therefore  more 
nearly  represented  by 

Fe2O3  +  6CO  =  2Fe  +  3CO  +  3CO2, 

which  shows  that  112  parts  of  iron  would  require  at  least  72 
parts  of  carbon  to  be  oxidized  at  the  tuyeres  to  CO,  in  order  to 
produce  the  gas  necessary  for  its  reduction.  The  presence  of 
some  CO2  in  the  furnace  coming  directly  from  carbonates  in 
the  charge  would  neutralize  still  more  of  the  reducing  power 
of  the  CO  gas,  and  cause  still  more  of  it  to  be  theoretically  re- 
quired for  reduction.  The  minimum  amount  of  carbon  neces- 
sary to  be  charged  in  the  furnace  will  be  that  necessary  to 
furnish  fixed  carbon  enough  for  this  reducing  gas  and  for  the 
carbon  in  the  pig  iron.  This  would  be,  per  100  parts  of  pig 
iron,  containing  say  93  iron  and  3  carbon,  and  using  coke  con- 
taining 90  per  cent,  of  fixed  carbon: 

Carbon  burnt  at  tuyeres  =  72X93-^112  =59.8 

Carbon  in  pig  iron  .       =3.0 

Total  fixed  carbon  necessary  =  62 . 8 

Total  coke  to  supply  this  =  62.8  4- 0.9  =69.8 

It  results  from  these  calculations  that  if  "  Griiner's  ideal 
working  "  of  a  blast  furnace  were  carried  out  to  the  practical 
extent  of  reducing  all  the  charge  by  carbonous  oxide,  CO, 
and  oxidizing  no  carbon  at  all  directly  above  the  tuyeres,  that 
about  63  parts  of  fixed  carbon  would  be  required  per  100  of 
pig  iron  made,  requiring  from  70  to  80  parts  of  fuel,  according 
to  its  richness  in  fixed  carbon  (90  to  80  per  cent.).  In  prac- 
tice, as  is  well  known,  more  than  this  is  commonly  used,  be- 
cause of  the  larger  proportion  of  unused  CO  in  the  gases  than 
above  assumed:  and  less  than  this  has  been  regularly  used, 
showing  that  economy  of  fuel  can  be  attained  without  adher- 
ing to  "  Griiner's  ideal  working,"  in  fact,  by  transgressing  it  as 
far  as  one  dares. 


UTILIZATION  OF  FUEL  IN  BLAST  FURNACE.          279 

The  principle  involved  can  be  best  grasped  by  a  calculation 
of  the  amount  of  carbon  which  would  be  required  by  the  fur- 
nace, supposing  all  the  heat  necessary  for  melting  down  the 
charge  were  supplied  by  electrical  means,  thus  dispensing,  for 
the  purposes  of  this  supposition,  with  the  necessity  of  blast 
and  the  consequent  necessity  of  oxidizing  any  carbon  by  any 
other  agent  than  the  oxygen  given  up  by  the  ore.  -  In  this  case 
the  gases  resulting  would  be,  let  us  assume,  of  the  same  com- 
position as  before,  that  is,  containing  equal  volumes  of  CO 
and  CO2,  and  since  this  oxygen  is  abstracted  altogether  from 
the  ore,  the  reaction  is 

• 

Fe203  +  2C  =  2Fe  +  CO  +  C02 

This  would  represent  the  utilization  of  carbon  in  an  electrically- 
heated  furnace,  and  would  require  per  100  of  pig  iron  made, 
assuming  it  3  per  cent,  carbon  and  93  iron : 

Carbon  for  reduction  24  X  93  -^  1 12  =19.9 

Carbon  in  pig  iron  =     3.0 

Total  fixed  carbon  necessary  =  22 . 9 

Or  only  a  little  over  one-third  as  much  as  the  minimum  re- 
quired when  the  smelting  down  is  done  by  blast. 

Aside  from  electrical  furnace  practice,  however,  this  discus- 
sion proves  that  whatever  fixed  carbon  burns  or  oxidizes  above 
the  region  of  the  tuyeres,  in  a  blast  furnace,  absorbs  oxygen 
from  the  charges  with  three  times  the  efficiency  of  carbon  first 
burnt  at  the  tuyeres.  Every  pound  of  oxygen  abstracted  from 
the  charges  by  solid  carbon  requires  the  use  or  intervention  of 
only  one-third  as  much  carbon  as  that  which  is  abstracted  by 
CO  gas;  or,  each  pound  of  carbon  abstracting  oxygen  directly 
from  the  charge  takes  from  it  three  times  as  much  oxygen  as 
a  pound  of  carbon  first  burnt  to  CO  at  the  tuyeres  possibly  can. 

The  ordinary  furnace  produces  at  the  tuyeres,  in  order  to  get 
heat  enough  to  melt  down  the  charges,  more  CO  gas  than  is 
needed  to  abstract  all  the  oxygen  from  the  charges;  under 
these  conditions  it  is  uneconomical  to  oxidize  any  carbon  at 
all  above  the  tuyeres.  The  exceptional  furnace,  because  of 
pure  ores,  small  amount  of  slag,  pure  fuel,  high  temperature 
of  blast,  or  dry  blast,  gets  heat  enough  at  the  tuyeres  to  melt 


280  METALLURGICAL  CALCULATIONS. 

down  the  charges  without  producing  enough  CO  gas  to  reduce 
all  the  charges;  under  these  conditions  more  or  less  reduction 
is  effected  by  solid  carbon,  and  with  the  greatest  economy  in 
quantity  of  carbon  required  in  the  furnace.  These  are  the 
conditions  under  which,  having  passed  the  turning  point,  the 
greater  economy  of  fuel  is  attained  the  farther  away  one  can 
get  from  "  Gruner's  ideal  working." 


CHAPTER  IV. 
THE  HEAT  BALANCE  SHEET  OF  THE  BLAST  FURNACE. 

Twenty-eight  years  ago,  Sir  Lothian  Bell  first  constructed 
a  satisfactory  heat -balance  sheet  for  a  blast  furnace.  His  ob- 
servations were  largely,  and  his  experience  altogether,  confined 
to  the  reduction  of  the  argillaceous  siderite  ores  of  the  Cleve- 
land district,  England,  and  although  he  made  numerous  at- 
tempts to  draw  general  conclusions  from  the  data  at  hand 
applicable  to  iron  smelting  in  general,  yet  many  of  his  deduc- 
tions remain  true  only  for  the  particular  ores  and  manner  of 
working  characteristic  of  the  Cleveland  district. 

No  treatment  of  this  subject,  however,  can  be  based  other- 
wise than  upon  Bell's  researches,  following  the  lines  laid  down, 
in  his  "  Principles  of  the  Manufacture  of  Iron  and  Steel." 

HEAT  RECEIVED  AND  DEVELOPED. 
The  items  on  this  side  of  the  balance  sheet  are: 

(1)  Combustion  of  carbon  to  carbonous  oxide  (CO). 

(2)  Combustion  of  carbon  to  carbonic  oxide  (CO2). 

(3)  Sensible  heat  of  the  hot  blast. 

(4)  Heat  of  formation  of  the  pig  iron  from  its  constituents. 

(5)  Heat  of  formation  of  slag  from  its  oxide  constituents. 
(1)  and  (2)   Combustion  of  carbon  in  the  furnace.     There  is 

but  one  satisfactory  way  to  determine  with  exactness  the 
amounts  under  this  heading.  From  the  balance  sheet,  the  total 
amount  of  carbon  passing  into  the  gases  is  obtained;  from  the 
analysis  of  the  gases,  the  weight  of  carbon  per  unit  volume 
of  gases  is  calculated;  the  first  divided  by  the  second  gives  the 
volume  of  gases  per  unit  weight  of  pig  iron  produced.  The 
amount  of  CO  and  CO2  in  these  gases  is  then  obtained  by  use 
of  the  gas  analysis,  and  if  from  the  total  CO  and  CO2  in  the 
gases  there  be  subtracted  the  CO  and  CO2  contributed  as  such 
by  the  solid  charges,  the  difference  is  the  CO  and  CO2  which 
have  been  formed  in  the  furnace.  The  heat  evolved  in  the 
formation  of  these  quantities  can  then  be  calculated. 

281 


282  METALLURGICAL  CALCULATIONS. 

Illustration. — In  Problem  51  it  was  calculated  that  per  1,000 
kilos,  of  pig  iron  produced,  534.09  kg.  of  carbon  went  into  the 
gases;  also  that  the  analysis  of  the  gases  showed  0.20736  kg. 
of  carbon  in  each  cubic  meter  of  gas.  The  quotient  indicated, 
therefore,  2575.6  cubic  meters  of  gas  produced  per  ton  of  pig 
iron.  From  the  analysis  of  the  gases  there  was  in  this  volume, 

2575.6X0.231  =  595.0  m3  of  CO 
2575.6X0.148  =  381.2  m3  of  CO2 

whose  weights  were 

595.0X1.26    =  749.7kg.  CO 
381.2X1.98    =  754.8kg.  CO2 

The  balance  sheet  shows,  however,  49.1  kg.  of  CO2  contained  in  the 
limestone  flux  used,  which  can  be  assumed  as  entering  the 
gases  bodily.  Subtracting  this  we  have  705.7  kg.  of  CO2  formed 
in  the  furnace,  and  749.7  kg.  of  CO,  containing  respectively 

705. 7 X       =  192.5  kg.  of  C  in  CO2 


749.7X^1  =  321.3  kg.  of  C  in  CO 


The  heat  generated  in  the  furnace  by  the  oxidation  of  carbon 
is,  therefore, 

192.5X8100  =  1,559,250  Calories 
321.3X2430  =     780,760       " 


2,340,010 

If  this  carbon  could  have  been  entirely  burnt  to  CO2,  there 
would  have  been  generated 

513.8X8100  =  4,161,780  Calories 

Showing  that  only  56  per  cent,  of  the  calorific  power  of  the 
carbon  was  developed  in  the  furnace;  the  other  46  per  cent, 
exists  as  potential  calorific  power  in  the  waste  gases,  and  part 


THE  HEAT  BALANCE  SHEET.  283 

of  it  is   really  put  back  into  the  furnace  as  sensible  heat  in 
the  hot  blast. 

There  is  a  little  doubt  as  to  how  to  consider  the  CH4  in  the 
gases;  that  is,  whether  the  heat  of  its  formation  should  be 
reckoned  in  as  developed  in  the  furnace.  This  would  be  (C,H4) 
=  22,250,  or  1,854  Calories  per  kg.  of  carbon  contained  therein. 
Its  presence  in  the  gas  probably  results  largely  from  the  dis- 
tillation of  the  fuel  at  a  high  temperature,  and  the  heat  re- 
quired to  disunite  the  CH4  from  the  solid  fuel  is  probably  as 
great  as  is  represented  by  its  heat  of  formation  from  carbon  and 
hydrogen.  The  item  is,  therefore,  a  doubtful  one,  and  as  far 
as  we  know,  we  may  be  coming  about  as  near  to  the  truth  by 
omitting  it  altogether  as  by  counting  it  in.  If  we  wished  to 
add  it  in  the  illustration  just  given  the  calculation  would  be: 

Volume  of  CH4  =  2575.6  X  0.005  =  12.88  m3 

Weight  of  C  =       12.88X0.54    =     6.9    kg. 

Heat  of  formation  6.9   X  1,854  =  12,793  cal. 

It  should  be  emphasized  that  in  this  calculated  heat  of  oxi- 
dation of  carbon  in  the  furnace,  no  account  has  been  taken 
whatever  of  where  in  the  furnace  this  heat  is  generated.  Above 
all,  the  mistake  should  not  be  made  of  supposing  that  the 
780,760  Calories  produced  by  formation  of  CO  represents  the 
heat  generation  at  the  region  of  the  tuyeres ;  nothing  could  be 
further  from  the  truth.  A  great  deal  of  carbon  is  burnt  to  CO 
at  the  tuyeres,  and  some  above  the  tuyeres,  but  a  goodly  pro- 
portion of  this  CO  oxidizes  by  abstracting  oxygen  from  the 
charge  and  becomes  CO2.  It  would  not  be  incorrect,  however, 
to  divide  the  heat  of  oxidation  of  carbon  in  the  furnace  into 
two  parts,  viz:  to  assume  all  the  carbon  as  first  forming  CO, 
and  part  of  this  CO  afterwards  forming  CO2,  corresponding  to 
the  amount  of  the  latter  formed  in  the  furnace.  If  this  were 
done,  we  would  have 

513.8  kg.  C  to  CO      =  513.8X2430    =  1,248,535  Cal. 
449.2  kg.  CO  to  CO2  =  449.2  X  2430  =  1,091,475    " 


2,340,010 

This  analysis  gives  us  the  information  that  of  the  total  heat 
generated  by  the  oxidation  of  carbon  in  the  furnace,  some- 


284  METALLURGICAL  CALCULATIONS. 

where  about  one-half  is  generated  by  its  burning  to  CO,  and 
the  other  half  by  the  further  oxidation  of  CO  to  CO2;  and  we 
also  know  that  the  larger  part  of  the  former  takes  place  at  the 
tuyeres,  and  all  of  the  latter  takes  place  during  the  reduction 
of  the  charges  in  the  upper  part  of  the  furnace. 

If  we  know,  however,  or  have  calculated  the  amount  of  the 
blast  received  by  the  furnace,  or,  more  properly  speaking,  the 
amount  of  oxygen  in  the  blast,  then  the  heat  generated  by 
oxidation  of  carbon  at  the  tuyeres  becomes  known.  In  the 
previous  illustration,  taken  from  Problem  51,  we  can  also  take 
from  the  same  problem  the  weight  of  oxygen  in  the  blast, 
557.7  kilos.  This  would  burn  557.7X0.75  =  418.3  kg.  of 
carbon  to  CO  at  the  tuyers,  generating  there 

418.3X2430  =  1,016,410  Calories, 

or  44  per  cent,  of  all  the  heat  generated  by  oxidation  of  carbon 
in  the  furnace,  leaving  1,323,610  Calories  as  generated  above  the 
tuyeres  by  the  agency  of  the  oxygen  of  the  charges.  These 
figures  tell  us  just  where  and  how  the  principal  items  of  heat 
were  generated  in  this  particular  furnace,  and  the  similar  cal- 
culation may  be  made  for  any  blast  furnace  for  which  we  have 
the  necessary  data. 

(3)  Sensible  heat  in  the  hot  blast.  To  calculate  this  item, 
we  need  to  know  the  weight  or  volume  of  the  different  con- 
stituents of  the  blast  and  their  temperature.  The  question  at 
once  arises,  as  to  what  base  line  of  temperature  shall  be  chosen. 
It  is  most -convenient  to  choose  0°  C.,  since  that  is  not  over 
15°  from  the  average  temperature  in  the  largest  iron  produc- 
ing countries.  However,  any  other  prevailing  temperature 
may  be  taken  as  the  base  line,  involving  merely  a  little  more 
calculation,  since  our  specific  heats  are  reckoned  from  0°  C. 
The  temperature  ought,  moreover,  to  be  taken  as  near  to  the 
tuyeres  as  possible,  to  properly  take  into  account  the  effect 
of  cooling  of  the  blast  in  the  bustle  and  feeder  pipes,  from 
radiation  and  expansion.  The  blast  consists  of  air  p  oper 
and  moisture,  the  former  with  a  mean  specific  heat  between 
0°  and  t°  C  of  0.303 +  0.000027t,  in  kilogram  Calories  per  cubic 
meter,  or  in  ounce  calories  per  cubic  foot,  the  latter  with  a 
similar  mean  specific  heat  of  0.34+  O.OOOlSt.  Since  the  mois- 
ture at  times  amounts  to  as  much  as  5  per  cent,  of  the  blast, 
it  should  be  calculated  separately. 


THE  HEAT  BALANCE  SHEET.  285 

Illustration:  With  the  outside  air  at  30°  C.,  and  saturated 
with  moisture  (raining),  calculate  the  heat  carried  into  a  blast 
furnace  by  blast  carrying  in  1859.1  kilos,  of  nitrogen,  the  tem- 
perature of  the  heated  blast  being  600°  C.  Barometer  720 
millimeters  of  mercury.  Temperature  base  line  0°  C. 

Solution:  One  cubic  meter  of  the  moist  blast,  as  taken  into 
the  blowing  cylinders,  carries  all  the  moisture  it  can  hold,  the 
tension  of  which  is  therefore  31.5  millimeters.  The  tension  of 
the  air  proper  present  is  therefore  720-31.5  =  688.5  milli- 
meters, and  each  cubic  meter  of  moist  air  carries 

31  5 

•    '  •  =  0.0438  cubic  meter  of  moisture,  and 

£»oo   fr 

'      =  0.9562  cubic  meter  of  air  proper. 
7 20 

Whatever  the  temperature  of  the  blast,  the  moisture  and  air 
proper  will  be  in  this  same  proportion  whenever  its  tempera- 
ture is  over  30°  C.  If  the  temperature  were  0°  C.  the  moisture 
would  be  mostly  condensed,  but  for  the  purposes  of  calculating 
the  heat  brought  in  we  may  assume  the  moist  air  to  be  at  0° 
C.,  with  its  moisture  uncondensed.  That  volume  of  blast 
which  would  be  1  cubic  meter  at  0°  and  760  mm.  pressure, 
would,  therefore,  bring  in,  at  600°  C.,  the  following  quantity 
of  heat : 

H20    0.0438X[0.34  +0.00015  (600)]    X600  =    11.3  Calories 
Air      0.9562 X [0.303 +  0.000027  (600)]    X  600  =  179.7      " 


Total         191.0 

Since  the  nitrogen  present  in  this  is 

0.9562  X  1.293  X-^  =  0.9511  kg. 
lo 

the  heat  brought  in  per  1859.1  kg.  of  nitrogen  is 

373>344  Calories. 


An  amount  equal  to  over  one-third  of  all  the  heat  generated 
by  combustion  of  carbon  at  the  tuyeres. 

(4)  Heat  of  formation  of  pig  iron  from  its  constituents.    The 


286  METALLURGICAL  CALCULATIONS. 

pig  iron  contains  several  per  cent.,  some  5  to  10  altogether, 
of  carbon,  silicon,  manganese,  phosphorus,  sulphur  and  other 
elements.  The  energy  of  their  combination  with  the  iron  is  a 
somewhat  indefinite  quantity,  and  in  no  case  can  be  consid- 
able.  Berthelot  states  the  energy  of  combination  of  carbon 
with  iron  as  (Fe3,C)  =  8,460,  which  would  be  705  Calories  per 
kilogram  of  carbon,  and  another  investigator  (Ponthiere) 
states  the  heat  of  combination  of  phosphorus  with  iron  to  be 
zero.  In  the  present  state  of  uncertainty  it  is  hardly  allowable 
to  add  in  any  other  than  the  heat  of  combination  of  the  car- 
bon in  the  iron,  and  leave  out  that  of  the  other  elements. 

(5)  Heat  of  formation  of  the  slag  from  its  constituent  oxides, 
Here  we  touch  upon  a  qur'itity  of  more  than  insignificant 
proportions,  yet  which  is  not  yet  quantitatively  known  with 
satisfactory  accuracy.  The  main  constituents  of  the  slag 
are  SiO2,  APO3,  CaO,  MgO  and  CaS,  which  are  provided  by 
clay,  limestone  and  iron  sulphide.  If  we  allow,  on  the  other 
side  of  the  balance  sheet,  for  the  heat  necessary  to  de-hydrate 
clay,  drive  carbonic  acid  off  carbonates,  and  break  up  iron  sul- 
phide and  enough  CaO  to  furnish  Ca  for  CaS,  we  are  then  en- 
titled, on  the  other  hand,  to  place  in  the  heat  evolution  column 
the  heat  of  combination  of  aluminum  silicate  with  lime  and 
magnesia,  the  heat  of  formation  of  CaS  and  its  heat  of  so- 
lution in  the  silicate  slag.  The  heat  of  formation  of  CaS  is 
94,300  Calories,  or  2,947  Calories  per  kilogram  of  sulphur; 
its  heat  of  combination  with  a  silicate  slag  is  unknown.  The 
heat  of  combination  of  lime  with  aluminum  silicate  has  been 
determined  only  for  the  proportions  3CaO  to  APSi2O7,  that  is, 
for  168  parts  of  CaO  uniting  with  222  parts  of  aluminum  sili- 
cate. This  has  been  determined  in  Le  Chatelier's  laboratory  as 
(3CaO,  Al2Si2O7)  =  33,500  Calories  which  is  200  Calories  per 
unit  of  CaO  combining,  or  150  Calories  per  unit  weight  of  A12O3 
+  SiO2.  The  calculation  would  be  made  on  the  basis  of  the 
amount  of  lime  (plus  lime  equivalent  of  magnesia  present), 
if  it  were  present  in  a  smaller  ratio  than  168  to  222  of  silica 
and  alumina,  and  on  the  basis  of  the  silica  and  alumina,  if 
their  ratio  to  the  summated  lime  were  less  than  222  to  168. 
It  is  probable  that  in  the  near  future  these  quantities  will  be 
known  more  accurately. 

One  item  of  heat  received  by  the  furnace  has  not  been  men- 


THE  HEAT  BALANCE  SHEET.  287 

tioned,  because  of  its  usual  absence,  viz.:  heat  in  hot  charges. 
Very  rarely  roasted  ore  comes  hot  to  the  furnace,  in  which 
case  its  sensible  heat  must  be  counted  in,  else  the  thermal 
sheet  of  the  furnace  will  be  that  much  out  of  balance. 

HEAT  ABSORPTION  AND  DISBURSEMENT. 

The  items  on  this  side  of  the  balance  sheet  are: 

(1)  Sensible  heat  in  waste  gases,  including  water  vapor  only 
as  vapor. 

(2)  Sensible  heat  in  outflowing  slag. 

(3)  Sensible  heat  in  outflowing  pig  iron. 

(4)  Heat  conducted  to  the  ground. 

(5)  Heat  conducted  and  radiated  to  the  air. 

(6)  Heat  abstracted  by  cooling  water,  tuyeres,  etc. 

(7)  Heat  for  de-hydrating  the  charges. 

(8)  Heat  for  vaporizing  water  from  charges. 

(9)  Heat  absorbed  by  decomposition  of  carbonates. 

(10)  Heat  absorbed  in  reduction  of  iron  oxides. 

(11)  Heat  .absorbed  in  reduction  of  other  metallic  oxides. 

(12)  Heat   absorbed  by   decomposition   of  moisture   of  the 
blast. 

(1)  Sensible  heat  in  waste  gases.  The  amount  of  these  gases 
is  known  only  from  the  carbon  contained  in  unit  volume,  by 
analysis,  and  the  known  weight  of  carbon  entering  and  leav- 
ing the  furnace.  If  there  is  much  fine  coke  carried  over  by 
the  blast,  allowance  must  be  made  for  the  carbon  in  it,  be- 
cause this  would  not  be  represented  in  the  gas  analysis.  The 
analysis  of  completely  dried  gas  is  that  usually  obtained,  be- 
cause if  the  gas  is  measured  without  drying,  an  uncertain 
amount  of  moisture  is  condensed,  and,  therefore,  it  is  usual 
to  dry  before  measuring  and  analyzing.  The  amount  of  mois- 
ture in  the  gases  is  either  assumed  as  that  driven  off  from  the 
charges,  as  shown  by  the  balance  sheet,  or  else  is  determined 
directly  by  drawing  the  gases  through  a  calcium  chloride  tube 
or  other  dessicating  apparatus.  Several  tests  should  be  made 
to  get  a  fair  average,  because  much  more  will  be  in  the  gases 
immediately  after  charging  than  immediately  before.  Dust 
must  be  excluded  from  the  drying  tube  by  filtering  the  gases 
through  dry  asbestos.  The  average  temperature  of  the  gases 
should  be  known  over  a  considerable  period,  a  thermo-couple 


288  METALLURGICAL  CALCULATIONS. 

in  the  down-comer  gives  this  most  accurately  and  more  uni- 
formly than  if  inserted  above  the  stock  line  in  the  furnace. 

The  weight  of  moisture  per  unit  volume  of  dry  gas  is  then 
converted  into  volume  at  standard  conditions  by  dividing  by 
0.81  (Icubic  meter  =  0.81  kg;  1  cubic  foot  =  0.81  ounce 
avoirdup.).  The  sensible  heat  of  the  gases  is  then  calculated, 
using  0°  C.  as  the  base  line,  and  the  proper  mean  specific  heats 
of  the  gases  per  unit  of  volume.  The  water  vapor  will  here 
be  considered  simply  as  a  gas,  and  its  sensible  heat  above  water 
vapor  at  0°  only  calculated.  This  leaves  the  latent  heat  of 
vaporization  of  this  water  to  be  considered  as  a  separate  item 
(606.5  Calories),  that  is,  as  heat  absorbed  by  reactions  in  the 
furnace,  thus  putting  it  on  exactly  the  same  footing  as  the 
CO2  in  the  gases  which  has  been  expelled  from  carbonates  in 
the  furnace.  By  so  proceeding  much  uncertainty  as  to  the 
heat  in  the  water  vapor  is  avoided. 

If  the  amount  of  flue  dust  is  considerable  its  quantity  should 
be  ascertained,  and  the  heat  in  it  also  calculated  and  added  in 
to  the  heat  in  the  moist  gases.  Its  specific  heat  may  be  ap- 
proximated as  so  much  carbon,  iron  oxide  and  silica,  the  pro- 
portions of  each  of  these  present  being  known. 

(2)  Sensible  heat  in  outflowing  slag.  The  weight  of  slag 
produced  is  seldom  taken  directly,  but  can  be  reckoned  up 
with  all  needful  accuracy  from  the  balance  sheet  of  materials 
entering  and  leaving.  Its  temperature  and  specific  heat,  solid 
and  liquid,  melting  point  and  latent  heat  of  fusion,  are  un- 
fortunately almost  always  unknown  factors.  The  one  datum 
which  is  needful,  however,  is  the  total  heat  in  a  unit  weight  of 
liquid  slag  as  it  flows  from  the  furnace,  and  this  is  not  a  diffi- 
cult quantity  to  obtain.  A  rough  calorimeter  with  a  reliable 
thermometer  and  containing  a  carefully  weighed  quantity  of 
water,  may  be  easily  constructed.  Some  liquid  slag  is  run 
directly  into  it,  and  by  observing  the  rise  of  temperature  and 
afterwards  filtering  out,  drying  and  weighing  the  granulated 
slag,  a  satisfactory  determination  can  be  arrived  at.  This  is 
corrected  to  0°  C.  by  using  an  approximate  specific  heat,  say 
0.20,  for  the  range  of  final  calorimeter  temperature  to  zero. 
In  this  connection  it  is  important  to  note  that  the  calorimetric 
determinations  of  Akerman  on  blast-furnace  slags,  give  the 
heat  in  the  just-melted  slag,  whereas  slags  flowing  out  of  a 


THE  HEAT  BALANCE  SHEET.  289 

furnace  are  considerably,  some  200°  to  500°  C.,  above  their 
melting  point,  and  therefore  contain  some  50  to  150  Calories 
more  heat  than  that  given  by  Akerman  for  a  slag  of  similar  com- 
position. Since  Akerman 's  values  run  from  350  to  400  Calories, 
the  actual  heat  in  the  outflowing  slag  may  be  between  400 
and  550  Calories.  Akerman  himself  states  that  an  average 
of  twenty-seven  Swedish  furnaces  gave  530  Calories  as  the 
actual  heat  in  unit  weight  of  outflowing  slag,  and  Bell  uses 
550  in  most  of  his  calculations  on  Cleveland  (England)  furnaces. 

(3)  Heat   in   outflowing  pig   iron.     The   heat   in   just-melted 
pig  iron  is  evidently  too  small  a  quantity  to  use  in  this  con- 
nection.    The  heat  in  the  outflowing  pig  iron  at  200°  to  500° 
above  its  melting  point  will  be  50  to  100  Calories  greater.     The 
former  quantity  is  about  245  Calories;  the  latter  will  be  300 
to  350.     Bell  takes  330  for  Cleveland  furnaces;  Akerman  states 
250   to   325   for   Swedish   furnaces.     We   may   conclude  then, 
to  use  300  Calories  for  a  coke  furnace  running  cool,  and  up 
to  350  Calories  for  a  very  hot  furnace. 

(4)  Heat  conducted  to  the  ground.     This  is  a  very  uncertain 
quantity.     It   varies  with  the  kind  of  ground,   and  is  more 
nearly  a  constant  per  day  than  per  unit  of  pig  iron  produced. 
It  is,   therefore,  expressed  per  unit  of  iron  produced,  larger 
for  small  furnaces  run  slowly  than  for  large  furnaces  run  fast. 
It  is  less  when  running  rich  ores  and  greater  with  poor  ores, 
other  things  being  equal.     As  nearly  as  can  be  assumed  we 
would  put  this  item  as  lying  between  60  and  200  Calories  per 
unit  of  pig  iron  made.     Bell  uses  169  on  one  Cleveland  furnace, 
but  it  is  certainly  less  than  100  in  some  charcoal  furnaces  using 
pure  ores  and  fuel,  and  consequently  with  a  small  heat  re- 
quirement. 

(5)  Heat   conducted   and  radiated   to   the   air.     This   item  is 
likewise  more  nearly  a  constant  quantity  per  day  for  a  given 
furnace,  and  is  therefore  less  per  unit  of  iron  produced  the 
faster  the  furnace  is  run.     It  may  vary  between  60  and  250 
Calories  per  unit  of  pig  iron,  the  former  in  furnaces  of  low  heat 
requirement   per   unit   of   iron   produced,   the   latter   in   those 
of  high  heat  requirement.     If  the  amount  were  calculated  it 
would  figure  out  as  a  time  function,  and  would  require  the 
temperature  of  the  outside  shell,  that  of  the  air,  the  velocity 
of  the  wind,  and  the  total  outside  surface,  in  order  to  calculate 


290  METALLURGICAL  CALCULATIONS. 

by  the  principles  of  heat  radiation  and  conduction,  the  amount 
radiated  per  day.  No  one.  has  done  this  yet  for  any  one  fur- 
nace, and,  in  brief,  items  (4)  and  (5)  of  this  schedule  are  usually 
grouped  together  and  determined  simply  by  difference,  their 
sum  aggregating  from  100  to  500  Calories  per  unit  of  pig  iron, 
averaging  100  to  150  for  charcoal  furnaces  of  low  heat 
requirement,  200  to  450  for  Cleveland  furnaces  (Bell),  and 
300  to  500  for  large,  modern  furnaces  with  thin  walls  and  great 
height. 

(6)  Heat  abstracted  by  cooling  water.     In   the  old-fashioned 
heavy  masonry,  cold  blast  furnace,  this  item  was  zero.     With 
the  advent  of  hot  blast,  the  water  needed  for  cooling  the  tuyeres 
entered  as  a  heat  abstracting  factor.     It  is  greater  the  harder 
a  furnace  is  blown,  but  does  not  increase  proportionately  with 
the  output.     The  heat  lost  by  tuyere-cooling  water  may    be 
50  to  100  Calories  per  unit  of  pig  iron  made.     That  for  cooling 
of  bosh  plates  and  the  outside  of  the  crucible  in  modern  fur- 
naces, may  vary  all  the  way  up  to  200  Calories.     These  two 
items  are  very  large  in  a  modern  furnace,  but  are  necessary 
expenditures  of  heat  energy  in  order  to  preserve  the  lines  of 
the  furnace  during  fast  running.     For  any  particular  furnace 
they  may  be  determined  with  all  needful  accuracy  by  mea- 
suring the  amount  of  water  pumped  or  used  for  these  purposes 
and  its  temperature  before  and  after  using. 

(7)  and   (8)   Drying  and  de-hydrating  charges.     Water  goes 
into  the  furnace  as  moisture  and  as  combined  water  of  the 
charges.     To  convert  the  moisture,  such  as  is  evaporated  by  a 
current   of   moderately   warm   air,    into   vapor   requires    606.5 
Calories  per  unit  of  water.     This  allows  merely  for  its  vapor- 
ization in  the  furnace,  and  not  for  any  sensible  heat  which  it 
may  carry  out  of  the  furnace  at  the  temperature  of  the  waste 
gases.     This   latter   item  is   properly   considered   in   with   the 
sensible  heat   of  the  waste  gases.     The   common   practice  of 
saying  that  it  takes  637  Calories  to  evaporate  the  moisture  of 
the  charges  is  wrong,  because  this  amount  would  convert  water 
at  0°  to  vapor  at  100°,  and,  therefore,  would  include  part  of 
what  is  properly  the  sensible  heat  of  the  waste  gases.     On  the 
other  hand,  it  is  equally  wrong  to  say  that  this  heat  of  vapor- 
ization should  be  counted  in  as  sensible  heat  in  the  hot  gases; 
it  would  be  just  as  logical  or,  rather,  equally  illogical,  to  count 
the  latent  heat  of  vaporization  of  CO2  as  sensible  heat  in  the  gases. 


THE  HEAT  BALANCE  SHEET.  291 

To  drive  off  water  of  hydration  from  hydrated  minerals  in 
the  charge  requires  an  additional  amount  of  chemically-ab- 
sorbed heat.  As  far  as  is  known,  this  is  small  for  the  water 
driven  off  hydrated  iron  oxides,  so  small  as  to  be  a  doubtful 
quantity  and  safely  left  out ;  but  if  it  comes  from  clay  the  large 
amount  of  611  Calories  is  absorbed  in  merely  separating  it 
from  its  chemical  combination  (2H2O,  APSi2O7)  =  22,000  Calories, 
which  would  require  611  +  607  =  1,218  Calories  to  put  into 
the  state  of  vapor  each  unit  weight  of  water  entering  the  fur- 
nace chemically  combined  in  clay.  (This  does  not  concern 
the  ordinary  moisture  in  moist  clay,  expelled  at  100°  C.,  but 
only  the  combined  water  in  the  dry  clay).  Where  much  clay 
occurs  in  the  ores  this  quantity  becomes  important,  and  its 
amount  explains  some  of  the  difficulties  met  in  working  clayey 
charges,  particularly  since  a  large  part  of  this  chemically  com- 
bined water  is  expelled  only  at  a  red  heat,  and,  therefore, 
cools  greatly  the  hotter  zones  of  the  furnace. 

(9)  Decomposition  of  carbonates.  Raw  limestone,  or  dolo- 
mite, is  the  usual  flux  of  the  blast  furnace,  and  its  carbonic 
acid  is  evolved  at  temperatures  between  600°  and  800°.  Whether 
some  of  this  is  subsequently  decomposed  by  contact  with 
carbon  and  reduced  to  CO,  is  immaterial  to  the  balance  sheet, 
because  more  than  enough  CO2  escapes  from  the  furnace  to 
represent  the  CO2  of  the  flux,  and  we  charge  the  furnace  only 
with  the  formation  of  the  CO  and  CO2  actually  found  in  the 
gases,  less  the  CO2  from  fluxes.  Bell  charges  up  the  heat 
absorbed  also  in  the  assumed  reaction  CO2  +  C  =  2CO,  but 
this  is  an  error,  because  it  is  doubtful  how  much  of  the  CO2 
is  thus  decomposed,  and  the  question,  in  its  last  analysis, 
is  one  of  heat  distribution  in  the  furnace,  and  does  not  concern 
the  totals  of  heat  absorbed  or  evolved.  We  can,  therefore, 
omit  the  item  of  decomposition  of  this  CO2  (as  likewise,  and 
for  analogous  reasons,  the  heat  evolved  in  carbon  deposition  in 
the  upper  part  of  the  furnace— 2CO  =  C  +  CO2),  and  need 
consider  only  the  heat  required  to  expel  the  CO2  from  car- 
bonates. This  is : 

(CaO,  CO2)    =  45,150  Calories  =  1,026  Calories   per  kg.  CO2 

(MgO,  CO2)  =  29,300       "         =      666 

(MnO,  CO2)  =  22,200       "         =      500       " 

(FeO,  CO2)    =  24,900       "        =      566       " 

(ZnO,  CO2)    =  15,500       "         =      352       " 


292  METALLURGICAL  CALCULATIONS. 

By  using  the  above  figures,  in  connection  with  the  known 
composition  of  ore  and  fluxes,  the  heat  required  to  decompose 
carbonates  can  be  correctly  calculated. 

(10)  Reduction  of  iron  oxides.     The  heats   of   formation   of 
the  various  oxides  of  iron  are: 

(Fe,O)        =     65,700  Calories  =  1,173  Calories  per  kg.  iron 
(Fe3,O4)      =  270,800       "         =  1,671 
(Fe2,O3)     =  195,600       "         =  1,746 

And,  therefore,  just  these  quanities  of  heat  are  required  per 
unit  weight  of  iron  reduced  from  these  compounds.  If  the 
ore  is  a  carbonate  the  heat  absorbed  in  driving  of  CO2  from 
FeCO3  can  be  first  allowed  for,  and  then  the  heat  required 
for  reduction  of  the  FeO  calculated  on  the  weight  of  the  re- 
duced iron.  If  some  FeO  goes  into  the  slag  it  will  be  as  FeO, 
and  if  the  ore  was  Fe203,  or  Fe304,  the  reduction  of  unit  weight 
of  iron  from  the  state  of  Fe203,  or  Fe3O4,  to  the  state  of  FeO, 
absorbs  respectively  573  or  498  Calories,  as  may  be  readily 
deduced  from  the  heats  of  formation  of  the  three  oxides  con- 
cerned. If  FeS  is  present  its  heat  of  formation  is 

(Fe,  S)  =  24,000  Calories  =  429  Calories  per  kg.  Fe. 

If  the  iron  is  charged  partly  as  silicate,  such  as  mill  or  tap 
cinder,  an  additional  amount  of  heat  will  be  required  for  re- 
duction, equal  to  that  needed  to  separate  the  iron  oxides  from 
their  combination  with  silica.  The  heat  of  formation  of  the 
bi-silicate  slag  only  has  been  determined. 

(FeO,  SiO2)  =  8,900  Calories  =  148  Calories  per  kg.  SiO2. 

And  since  the  cinders  concerned  contain  relatively  more  iron 
than  this,  we  can  best  make  allowance  for  the  heat  required  to 
set  free  the  silica.  It  is  necessary,  therefore,  to  take  the  amount 
of  silica  in  the  iron  cinder  charged,  and  allow,  as  necessary 
for  its  decomposition  into  FeO  and  SiO2,  148  Calories  for  each 
unit  weight  of  SiO2  contained. 

(11)  Reduction    of    non-ferrous    oxides.     Silicon    is    usually 
present  in  pig  iron,  its  reduction  from  silica  requiring: 

(Si,  O2)  =  180,000  Calories  =  6,413  Calories  per  kg.  Si. 
There  is  a  little  doubt  about  this  (Berthelot's)  figure;  more 


THE  HEAT  BALANCE  SHEET  293 

recent  determinations,  not  yet  published,  point  rather  to  196,000 
and  7,000  Calories  respectively. 

Manganese  is  often  present  in  the  ores  as  Mn2O3,  Mn3O4,  or 
MnO2,  and  going  partly  into  the  slag  as  MnO.  The  heat  ab- 
sorbed in  reduction  to  manganese  is: 

(Mn,  O)        =    90,900  Calories  =  1,653  Calories  per  kg.Mn 
(Mn3,  O4)      =  328,000       "         =  1,988       " 
(Mn,  O2)      =  125,300       "         =  2,278       " 

For  the  MnO  produced  and  going  into  the  slag,  the  reduc- 
tion from  Mn3O4,  or  MnO2,  per  unit  weight  of  contained  man- 
ganese, absorbs  335  or  625  Calories  respectively. 

Sulphur  generally  comes  from  the  reduction  of  FeS,  re- 
quiring 667  Calories  per  kg.  of  sulphur;  but  care  must  be  taken 
not  to  allow  for  this  heat  twice,  since  if  reckoned  once  under 
the  head  of  iron  reduction  it  must  not  be  reckoned  on  the 
balance  sheet  a  second  time  under  sulphur.  One  reduction  of 
FeS  liberates  both  constituents. 

Phosphorus  may  be  reduced  in  large  quantity  in  making  basic 
iron.  It  probably  comes  mostly  from  calcium  phosphate,  in 
which  case  we  must  reckon  on  not  only  the  heat  of  oxidation 
of  phosphoric  oxide  but  also  its  heat  of  combination  with  lime: 

(P2,  O5)  =  365,300  Cal.  =  5,892  Cal.  per  kg.  of  P. 

(3CaO,  P2O5)    =  159,400    "     =  2,410    "         "         contained. 

Making  a  total  heat  requirement  of  8,302  Calories  to  separate 
unit  weight  of  phosphorus  from  phosphate  of  lime,  and  leave 
free  lime.  In  a  furnace  making  a  pig  iron  with  several  per 
cent,  of  phosphorus,  this  item  becomes  quite  large. 

Calcium  occurs  in  the  slag  as  CaS,  its  reduction  from  lime 
requiring 

(Ca,  O)  =  130,500  Calories  =  3,263  Calories  per  kg.  Ca. 

Other  elements  than  the  above  rarely  occur  in  pig  iron  in 
notable  quantity.  If  rare  ones  occur,  their  heat  of  reduction 
can  be  sought  in  thermochemical  tables.  Those  of  tungsten, 
titanium,  molybdendum  and  chromium  are,  however,  not  at 
present  known. 

(12)  Decomposition  of  moisture  in  the  blast.     This  is  to  be 


294  METALLURGICAL  CALCULATIONS. 

counted  as  vapor  of  water,  the  heat  required  to  decompose, 
which  is: 

(H2,  O)  vapor  =  58,060  Cal.  =     3,226  Cal.  per  kg.  H2O. 

=  29,030    "        "     H2. 

It  is  not  correct  to  allow  here  for  the  sensible  heat  in  this 
water  vapor  coming  in  with  the  hot  blast,  because  that  heat 
should  go  on  the  other  side  of  the  balance  sheet  as  heat  de- 
livered to  the  furnace.  Neither  is  it  correct  to  subtract  the 
heat  of  combination  of  the  oxygen  of  this  moisture  with  car- 
bon to  form  CO  at  the  tuyeres;  because,  although  that  com- 
bination actually  does  take  place,  yet  the  heat  thereby  evolved 
properly  belongs  also  on  the  other  side  of  the  balance  sheet, 
and  is  there  properly  taken  care  of  as  part  of  the  heat  of  oxida- 
tion of  carbon  in  the  furnace. 

Problem  57. 

(Data  partly  from  paper  by  the  author,  Transactions  Ameri- 
can Institute  of  Mining  Engineers,  1905). 

A  blast  furnace  running  on  Lake  Superior  ore  has  the  fol- 
lowing charges,  per  100  of  pig  iron  produced: 

Hematite  ore:  177.6;  composition              H2O  — 10.0  per  cent. 

SiO2  —10.0 

A12O3  -  3.5 

Fe203  —76.5 

Limestone:  44.4;  composition                    SiO2  —  5.0 

MgO  —  4.8 

CaO  -47.6 

CO2  —42.6 

Coke:  95.8;composition                               SiO2  -  5.3 

CaO  —  5.3 

H2O  -  1.0 

C  —88.0 

Pig  iron  produced:  100;  composition        Si  —  1.0 

C  —  4.0 

Fe  —95.0 

Gases  produced :  composition  dry :            CO2  — 13.0        "v»\. 

CO  —22.3 

N2  —64.7 


THE  HEAT  BALANCE  SHEET.  295 

Blast  used:  Contained  5.66  grains  of  moisture  per  cubic  foot 

of  dry  air,  at  24°  C.  =  75°  F. 

Charges  in  pounds:  Coke,  10,200;  ore,  20,000;  stone  5,000. 
Product  per  day:    Pig  iron,  358  tons  =    801,920  pounds. 
Coke  used  per  day:  768,626  pounds. 
Temperature  of  blast  720°  F.  =  382°  C. 
Temperature  of  waste  gases  =  538°  F.  =  281°  C. 
Displacement  of  blowing  engines  =  40,000  ft.3  per  min. 
Heat  in  one  unit  of  pig  iron  =  325  Calories. 
Heat  in  one  unit  of  slag  =  525  Calories. 
Cooling  water,  per  day,  heated  50°  C.  =  300,000  gallons. 

Required;     (1)  The  volume  of  gases  per  100  kg.  of  pig  iron 

(2)  A  balance  sheet   of  materials  entering  and  leaving  the 
furnace,  per  100  unit  of  pig  iron. 

(3)  The  volume  and  weight  of  blast  per  100  kg.  of  pig  iron. 

(4)  The  efficiency  of  the  blowing  plant. 

(5)  The  heat  balance  sheet  of  the  furnace. 

(6)  The  proportion  of  the  fixed  carbon  of  the  fuel  burnt  at 
the  tuyeres. 

(7)  The    proportion    of    the    whole    heat    generated  at  the 
tuyeres. 

(8)  The  proportion  of  the  iron  reduced  in  the  furnace  which 
is  reduced  by  solid  carbon  from  FeO. 

(9)  The  theoretical  maximum  of  temperature  at  the  tuyeres. 

(10)  The    theoretical    maximum    if    all    the    moisture    were 
removed  from  the  blast. 

Solution:     (1)    To  find  the  volume  of  the  gases  : 

Carbon  in  the  coke  used  95.8X0.88  =  84.3  kg. 

Carbon  m  the  limestone  44.4X0.426  X     -  =     5.2    " 


Total  carbon  entering  furnace  =89.5 

Carbon  in  100  of  pig  iron  =     4.0 

Carbon  entering  gases  =85.5 

Carbonous  and  carbonic  oxides  in  gases  =  35.3  per  cent 

Carbon  in  1  m3  of  dried  gas  =  0.54X0.353  =     0.19062  kg. 

85  5 

Dry  gas  per  100  kg.  of  pig  iron  =   ()  19Q62  =  448.5  m3.  (1) 


296  METALLURGICAL  CALCULATIONS. 

Dry  gas  per  100  oz.  of  pig  iron  =  448.5  ft.3 

Dry  gas  per  100  pounds  of  pig  iron  =  448.5 

X16  =7176.0" 

Dry  gas  per  2240  pounds  of  pig  iron  =  7176.0 

X22.4  =  160,742  " 

160,742X358  (tons) 
Dry  gas  per  minute  = '• — 6Q><24 =     39,962" 

(2)  BALANCE  SHEET  OF  MATERIALS,  PER  100  OF  PIG-IRON. 


CO 

{^ 

Charges 
Fe2O3  135.6 

Pig  Iron                    Slag 
Fe  95  .  0                  

Gases 

o 

40  7 

1-* 
o 

H2O       17.8 

H2O 

7  8 

6 

SiO2       17.8 

Si      1.0             SiO2   15.7 

O 

1.1 

•^ 

SiO2         2.2 

SiO2     2.2 

s 

CaO       21.1 

CaO   21.1 

q 

MgO         2  .  1 

MgO     2  .  1 

s 

CO2        19  0 

CO2 

19  0 

C            84  3* 

C      40 

c 

80  3 

% 

SiO2         5  .  3 

SiO2     5.3 

H 

CaO         5  .  3 

CaO     5.3 

fe 

H2O         0  9 

H2O 

0  9 

(N 

O2          96  .  4 

o 

96.4 

s 

N2        321  3 

N2 

321  3 

•s 

H2O         4.5 

1  H 

0  5 

5 

[o 

4.0 

Total   740.0 

100.0                   58.0 

582.0 

The  charges  are  calculated  simply  from  the  weights  of  ore, 
flux  and  fuel  used,  and  their  percentage  composition.  The 
blast  is  calculated  as  given  in  solution  of  requirement  (3). 

The  1.0    of   silicon    in    the   pig  iron  requires  1.0  X  (60-5-28) 
=  2.1  parts  of  SiO2  to    furnish  it,  leaving  15.7  of   unreduced 
SiO2  to  go  into  the  slag,  and  1.1  of  oxygen  to  the  gases. 

(3)  The  balance  sheet  shows  that  the  solid  charges  furnish 
to  the  gases  41.8  of  O,  190  of  CO2  and  18.7  of  H20.  The  H2O 
goes  as  such  into  the  gases,  and  therefore  its  oxygen  is  not 
present  in  the  sample  of  dry  gas  analyzed.  The  oxygen  in 
CO2  is  19.0  X  (32 -r- 44)  =  13.8  kg.,  which  added  to  the  41.8 
gives  55.6  of  oxygen  getting  into  the  gases  as  CO  or  CO2,  from 
the  solid  charges. 


THE  HEAT  BALANCE  SHEET.  297 

The  oxygen  in  the  CO  and  CO2  of  the  gases  is  to  be  calcu- 
lated from  the  oxygen  in  unit  volume  of  gas  and  the  total 
volume  of  gas  produced.  The  oxygen  in  1  cubic  meter  of  gas 
will  be: 

O  in  CO  =  0.223X    (o.09  X  y \    X  ^|  =  0.16056kg. 

(0.09  X  f ) 


roo 
0.09X  -^1    X  =  0.18720 


0.34776    ' 

A  quicker  solution  is  to  note  that  CO2  represents  O2,  its  o/m 
volume  of  oxygen,  and  CO  represents  O,  or  half  its  volume  of 
oxygen,  and  since  1  cubic  meter  of  oxygen  weighs  0.09X  (32-*  2) 
=  1.44  kilograms,  the  weight  of  oxygen  required  is 


(n  22*3  \ 

-^~  +  0.13J  Xl.44  =  0.34776  kg. 


The  total  oxygen  in  448.5  m3  of  gases  is  therefore 

448.5X0.34776  =  156     kg 

subtracting  that  furnished  by  the  solid  charges         =     55.6   " 
there  remains  oxygen  furnished  by  blast  =100.4    " 

If  the  blast  were  perfectly  dry  air,  its  nitrogen  would 

be  100.4  X  (10  *3)  =  334.7    " 

and  the  weight  of  dry  blast  =  435.1    " 

and  its  volume  at  0°  C.  435.1-5-1.293  =  336:5  in3 

The  blast,  however,  is  not  dry,  and  the  100.4  kilos,  of  oxygen 
includes  that  brought  in  by  the  moisture.  The  moisture  weighs 
5.66  grains  per  cubic  foot  of  measured  dried  air;  it  is,  there- 
fore, simplest  to  calculate  the  oxygen  entering  the  furnace  per 
unit  volume  of  air  proper  entering.  Since  5.66  grains  =  5.66 
^437.5  =  0.01294  ounces  av.,  the  calculation  can  be  made  in 
ounces  per  cubic  foot  or  kilograms  per  cubic  meter  in  identical 
figures  as  follows: 

Oxygen  in  0.01294  parts  of  water  =  0.01294  X-f-          =0.0115 

y 

Oxygen  in  1  volume  of  air  proper,  at  24°  C. 

o  070 

-1-283*  13*   278  +  24  =°'2743 


Sum  =  0.2858 


298  METALLURGICAL  CALCULATIONS. 

The  blast  received,  per  100.4  kg.  of  oxygen  thus  received,  is 
therefore,  measured  dry  at  24°  C: 

'  351.8  cubic  meters, 


n 

(J  . 

and  in  cubic  feet,  per  100.4  pounds  of  oxygen: 
100.4X16 


0.2858 


=  5628.8  cubic  feet. 


The  volume  of  the  moist  air  containing  this  will  be  the  volume 
of  assumed  dried  air  plus  the  volume  of  the  water  vapor.  The 
latter  is,  per  unit  volume  of  dried  air: 

0. 01294 -f-  (o.OQXy   X  27\^  24)   =0.01738 

and  the  volume  of  moist  air  received,  at  24°,  is,  therefore, 

351.8X1.01738  =     357.9  cubic  meters. 
Or  5628.8X1.01738  =  5726.6  cubic  feet. 

The  weights  of  water,  oxygen  and  nitrogen  received  per  100 
of  pig  iron  are  (as  already  given  on  the  balance  sheet): 

H20    0.01294X351.8=      4.52kg. 
O2       0.2743   X351.8  =     96.4      " 
N2       0.9143   X351.8  =  321.3     " 

And  the  volumes  of  these,  considered  theoretically  at  0°  C.  and 
standard  pressure,  per  100  kg.  of  pig  iron  made: 

H2O  vapor      4 . 52 -r-  0 . 81    =      5.6  cubic  meters. 
Air  417.7   -7-1.293  =  322.8     " 


Sum  =  328.4 

There  are  several  other  modifications  of  this  solution  which 
will  suggest  themselves  to  anyone  who  studies  out  the  question, 
but  while  half  a  dozen  ways  may  be  equally  correct,  that  one 
is  logically  to  be  preferred  which  is  most  easily  understood 
and  is  the  shortest.  One  solution,  however,  based  on  volume 
relations,  is  well  to  know.  Water  vapor,  H2O,  represents  half 


THE  HEAT  BALANCE  SHEET.  299 

its  volume  of  contained  oxygen,  while  air  has  0.208  oxygen. 
The  cubic  foot  of  dried  air  at  24°  C.  was  accompanied  by 

070  j_  04 
0.01294^-0.  SIX      *JI     •  =  0.0174  cubic  foot 


of   moisture,    which   was   removed   in   making   the   test.     The 
oxygen  per  cubic  foot  of  dry  blast  was,  therefore,  at  24°  C.  : 

0  as  H20  =  0.0174  -v-  2          =  0.0087  cubic  foot. 
O2  as  air     =  1.0000X0.208  =  0.2080      " 

0.2167      " 
Weight  of  this  oxygen  : 

070 


And  volume  of  dry  blast,  at  24°,  per  100  of  pig  iron: 
100.4 


0.2868 


350.1  cubic  meters. 


The  difference  of  about  0.4  per  cent  between  this  and  the  pre- 
viously-calculated result  is  due  to  the  figures  not  being  carried 
out  to  a  sufficient  number  of  decimal  places. 

(4)  The  efficiency  of   the   blowing  plant  is   found  by  calcu- 
lating the  volume  of  air  and  moisture  received  by  the  furnace 
per  minute,  calculated  to  24°  C.  =  75°  F.,  and  comparing  this 
with  the  piston  displacement  of  40,000  cubic  feet  per  minute. 

Volume  of  moist  air  received,  at  24°  C.,  per 

100  Ibs.  of  pig  iron  made  =  5726.6  ft.3 

Volume  per  day  =  5726.6X8019.2  =  45,922,750  ft.3 

Volume  per  minute  =  45,922,750-^1440       =         31,891  " 

o-j    CQf) 

Efficiency  of  blowing  plant  -  =79.7  per  cent.    (4) 

4U,UUU 

(5)  The  heat  balance  sheet  is  based  for  most  of  its  data  upon 
the  balance  sheet  of  materials,  the  calculations  already  made 
and  the  additional  data  given  in  the  statement. 

The  balance  sheet  shows  80.3  kilos,  of  carbon  oxidized  in 
the  furnace.  Of  this,  the  amount  oxidized  to  CO  and  remain- 


300  METALLURGICAL  CALCULATIONS. 

ing  as  such  in  the  gases  is  obtained  from  the  amount  of  CO  in 
the  gases,  as  follows: 

C  in  CO  =  448.5X0.223X0.54  =  54.0  kg. 

C  oxidized  to  CO2  is  therefore  80.3-  54  =  26.3  kg. 

The  heat  in  hot  blast  is  calculated  from  its  volume  assumed 
to  be  at  0°  C.,  and  already  calculated,  viz.;  322.8  cubic  meters 
of  air  proper  and  5.6  cubic  meters  of  water  vapor,  with  mean 
specific  heats  per  cubic  meter  between  0°  and  382°  C.  of  0.313 
and  0.40  respectively. 

The  heat  of  formation  of  the  pig  iron  may  be  taken  from 
the  amount  of  carbon  in  the  iron,  as  705  Calories  per  kilogram 
of  carbon,  and  that  of  silicon  omitted  from  consideration. 

The  heat  of  formation  of  the  slag,  since  it  contains  29.5  of 
silica  and  alumina  to  28.5  of  lime  and  magnesia,  may  be  taken 
as  150  Calories  per  unit  of  silica  and  alumina. 

The  total  heat  received  and  generated,  and  to  be  accounted 
for,  in  the  furnace,  is  therefore,  per  100  kg.  of  pig  iron: 

Carbon  to  CO   54.0X2430  =  131,220  Calories. 

Carbon  to  CO2  26.3X8100  =  213,030 

In  H2O  vapor     5.6X0.40   X382  \  on  QQK 

In  air  proper  322.8X0.313X382  / 

Solution  of  C  in  iron  4x705  =       2,820 

Formation  of  slag  29.5X150  =      4,425 


-t 


Total  =  390,880 

The  items  of  heat  distribution  are  325  Calories  in  each  kilo- 
gram of  pig  iron,  525  Calories  in  each  kilogram  of  slag,  heat 
in  the  waste  gases  at  281°,  heat  in  cooling  water,  lost  by  radia- 
tion and  conduction  (by  difference),  evaporation  of  the  mois- 
ture in  charges,  expulsion  of  CO2  from  carbonates,  decom- 
position of  moisture  of  blast,  reduction  of  silicon  and  iron. 

Reduction  of  Fe  95  kg.X  1,746     =  165,870  Calories. 

Reduction  of  Si  1  "  X  7,000      =       7,000 

Expulsion     of     CO2  from 

CaCO3  16.7  "  XI. 026 

Expulsion    of       CO2  from 

MgCO3  2.3  "  X     666 

Evaporation  of  H2O  18.7  "  X606.5     -     11,342 


THE  HEAT  BALANCE  SHEET.  301 


X281°  =     43,836 


Heat  in  waste  gases: 

Nand  CO    400  m3X  0.3106 
CO2  58.3  m3X  0.446 
H2O23.1  m3X  0.382 
Decomposition  of  moisture  of  blast: 

H2O  4.5  kg.  X  (29,040 -r9)  =     14,511 

Heat  in  slag  58  kg.  X525  =     30,450 

Heat  in  pig  iron  100X325  =     32,500 

Heat  in  cooling  water  (300,000  gallons  per 

diem)  300,000X8.3  (Ibs.)  X  50° -r- 8019.2  =     15,525 
Loss  by  radiation   and   conduction    (differ- 
ence) =    51,180 


Total  =  390,880        "  (5) 

(6)  The  proportion  of  the  fixed  carbon  burned  at  the  tuyeres 
is  obtained  by  calculating  the  weight  of  carbon  which  the 
oxygen  entering  at  the  tuyeres  could  oxidize  to  CO,  as  follows: 

Oxygen  entering  at  the  tuyeres  =  100.4  kg.  . 

Carbon  burned  to  CO  =  100.4X^1      =    75.3" 

lo 

Fixed  carbon  charged  =     84.3  ~" 

Proportion  burned  at  tuyeres  =    89.3  per  cent. 

A  more  just  proportion  is  that  between  the  carbon  burned 
at  the  tuyeres  and  the  total  fixed  carbon  oxidized,  because  the 
fixed  carbon  which  carbonizes  the  iron  cannot  be  oxidized  under 
any  circumstances.  This  proportion  in  this  furnace  is: 


indicating  a  very  fair  approximation  to  Gruner's  ideal  work- 
ing. If  we  make  the  further  allowance,  that  the  silicon  in  the 
pig  iron  is  necessarily  reduced  by  solid  carbon,  and  that  there- 
fore the  solid  carbon  needed  to  reduce  the  1  kilogram  of  silicon 
[IX  (24^-28)]  is  in  no  case  available  for  combustion  at  the 
tuyeres,  we  have  the  approach  to  Gruner's  ideal  working  meas- 
ured by  the  ratio 


80.3-0.9 


302  METALLURGICAL  CALCULATIONS. 

in  spite  of  which,  however,  the  furnace  is  not  doing  very  good 
work. 

(7)  The   proportion   of  the   heat   requirement   generated   or 
available  at  or  about  the  tuyeres  is  determined  as  follows: 

Combustion  of  C  to  CO  =  75.3X2430  =  182,979  Calories. 

Heat  in  hot  blast  =    39,385 

Format  on  of  pig  iron  =       2,820 

Formation  of  slag  =      4,425         ". 


Total  =  229,609 
Against  which  must  be  charged  heat  required 

to  decompose  moisture  of  blast  =     14,511 


Leaving  net  heat  available  =  215,098         " 

which  is  215,098^-390,880  =         55  per  cent, 

of  the  total  heat  generated  and  available  in  the  furnace. 

Another  way  in  which  it  is  sometimes  put,  is  that  the  oxida- 
tion of  carbon  to  CO  or  CO2  furnishes  a  certain  amount  of 
heat  to  the  furnace  (346,250  Calories  in  this  case),  of  which 
a  certain  amount  is  generated  by  combustion  at  the  tuyeres 
(182,979  Calories),  making  the  ratio  thus  considered,  in  this 
case,  53  per  cent., — which  is  almost  the  same  figures  as  above, 
but  not  so  significant,  since  it  is  illogical  to  consider  the  heat 
brought  in  by  the  hot  blast  as  not  being  available  heat  for 
doing  work  in  the  tuyere  region. 

(8)  The  proportion  of  iron  reduced  by  solid  carbon  is  found 
by  finding  how  much  carbon  is  used  up  in  that  reduction. 
Fixed  carbon  charged  =84.3  kg. 

Fixed  carbon  carbonizing  the  pig  iron          =     4.0    " 


Fixed  carbon  oxidized  in  the  furnace  =80.3 

Fixed  carbon  oxidized  by  the  blast  =  75.3 


Fixed  carbon  oxidized  above  the  tuyeres    =    5.0    " 
Carbon  needed  to  reduce  1  kg.  silicon          =    0.9   " 


Carbon  used  for  reducing  FeO  =    4.1 

P/> 
Amount  of  Fe,  thus  reduced  =  l.lX-r         =  19.1 


" 


Proportion  of  the  total  Fe,  thus  reduced 

-  ~  =20  per  cent.  (8) 


THE  HEAT  BALANCE  SHEET.  303 

(9)  The  maximum  temperature  of  the  gases  in  the  region 
of  the  tuyeres  is  that  temperature  to  which  the  heat  there 
available  will  raise  the  products  of  combustion.  This  question 
is  best  resolved  by  simply  considering  the  combustion  of  1 
kilogram  of  carbon,  evolving  2430  calories,  while  the  heat  in 
the  hot  carbon  itself  just  before  it  is  burnt,  and  that  in  the 
hot  blast  required,  will  also  exist  as  sensible  heat  in  the  pro- 
ducts,— the  whole  diminished  by  the  heat  necessary  to  decom- 
pose the  water  vapor  blown  in. 

Since,  per  75.3  kg.  of  carbon  burned  at  the  tuyeres  there 
enter  5.6  m3  of  water  vapor  and  322.8  m3  of  air  proper,  meas- 
ured at  0°,  the  volume  of  blast  per  kilogram  of  carbon  oxidized 
at  the  tuyeres  is 

H2O      5.6-V-75.3  =  0.0738m3  =  0.0598  kg. 
Air   322.8^75.3  =  4.2869m3 

The  products  of  the  combustion  are,  per  kg.  of  carbon  burned: 

CO  22.22-^12  =  1.8519  cubic  meters. 

N2  321.3*  1.26-v- 75.3  =3.3865     " 

H2  equal  to  H2O  decomposed  =  0.0738     " 


Total  =  5.3122 
The  heat  available  to  raise  their  temperature  is: 

Heat  of  combustion  of  1  kg.  carbon  =  2430  Calories. 

Heat  in  hot  blast  =  39,385^-75.3  =     523 

Heat  in  hot  carbon  at  t°  (or  very  nearly)  =  0.5t-feo  " 

Less    heat    absorbed    in    decomposing    steam 

14,511-*- 75.3  =     193 

Net  heat  available  in  gaseous  products  =2640  +  0.5t  Cal 

Calorific  capacity  of  gaseous  products 

=  5.3122  (0.303t  +  0.000027t2) 

Therefore        5.3122  (0.303t  +  0.000027t2)  =  2640  +  0.5t 
Whence  t  =  1910°  (9) 

This  represents  the  absolute  maximum  of  temperature  which 
the  gaseous  products  formed  at  the  tuyeres  can  possess. 

(10)  If  all  the  moisture  were  removed  from  the  blast,  the 
heat  available  would  be: 


304  METALLURGICAL  CALCULATIONS. 

By  combustion  of  1  kg.  carbon  =  2430  Calories. 

Heat  in  4.4685  m3  of  dry  air  at  382°  C. 

=  4.4685X0.313X382  =     574 

Heat  in  1  kg.  of  carbon  at  t°  =  0.5t-  120" 

Net  heat  available  in  gaseous  products  =  0.5t  +  2884  Cal. 

Calorific  capacity  of  gaseous  products 

=  5.3976  (0.303t  +  0.000027t2) 

Therefore         5.3976  (0.303t  +  0.000027t2)  =  2884  +  0.5t 
Whence  t  =  2018°  (10) 

It  is  to  be  noted  that  this  is  108°  C.  =  194°  F.  higher  than 
with  moist  blast ;  and  while  the  slag  and  iron  in  passing  through 
this  zone  of  high  temperature  will  not  reach  this  maximum 
temperature,  yet  they  would  be  heated  approximately  100°  C. 
higher  when  using  dry  blast,  if  all  other  conditions  were  kept 
constant. 

N.  B. — In  working  this  problem,  the  metric  measurements 
and  English  measures  have  been  purposely  used  interchange- 
ably, in  order  that  readers  may  understand  better  that  if  weights 
are  taken  in  pounds  and  the  heat  unit  is  the  pound  Calorie 
(1°  C.) ,  the  same  numbers  represent  a  solution  in  either  system. 
When  volumes  are  concerned,  cubic  feet  and  ounces,  or  ounce 
calories,  have  practically  the  same  numerical  expression  as 
cubic  meters  and  kilograms  or  kilogram  calories. 


CHAPTER  V. 
THE  RATIONALE  OF  HOT-BLAST  AND  DRY-BLAST. 

Problem  58  (for  practice). 

The  blast  furnace  of  Problem  57  had  its  blast  dried  before 
using,  to  the  extent  of  leaving  on  an  average  1.75  grains  of 
moisture  per  cubic  foot  of  air  proper,  at  —  5°  C.,  in  the  air 
pumped.  The  composition  of  the  ore,  limestone,  and  coke  used 
was  unchanged,  also  that  of  the  pig  iron  made.  The  weights 
charged  per  100  of  pig  iron  were:  Ore  177.6,  flux  44.4,  fuel  77.0, 
and  the  blast  used  calculates  out  oxygen  76.5,  nitrogen  255.0, 
moisture  1.0.  Analysis  of  gases:  CO  19.9,  CO2  16,  N2  64.1  per 
cent.  Product  per  day  447  tons.  Temperature  of  gases  191° 
C.,  of  blast  465°  C.  Piston  displacement  (air  at  -  5°  C.)  34,000 
cubic  feet  per  minute.  Assume  heat  in  pig  iron  and  slag  same 
as  before,  in  cooling  water  20  per  cent,  greater  per  day. 

Required. — (1)  The  volume  of  gases  per  100  kg.  of  pig  iron 
made. 

Answer. — Measured  dry,  355.9  m3. 

(2)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace,  per  100  units  of  pig  iron.     (See  table  on  page  281.) 

(3)  The  volume  of  blast  per  100  kg.  of  pig  iron. 
Answer.— 252.9  m3  at-  5°  C. 

(4)  The  efficiency  of  the  blowing  plant. 
Answer. — 82.3  percent. 

(5)  The  heat  balance  sheet  of  the  furnace,  per  100  of  pig  iron. 

Developed. 

Carbon  to  CO 92,950  Calories. 

^Carbon  to  CO2 206,955 

""  HeaV;.in  hot  blast 37,850 

Soluti^a;  of  carbon  in  iron 2,820         " 

Formation  of  slag 4,260 

Total 344,835 

305 


306  METALLURGICAL  CALCULATIONS. 

Distributed. 

Reduction  of  iron 165,870  Calorics. 

Reduction  of  silicon 7,000 

Expulsion  of  carbonic  oxide  (CO2) 18,666 

Evaporation  of  moisture  of  charges 11,342 

Heat  in  waste  gases 23,799         " 

Decomposition  of  moisture  of  blast 3,225 

Heat  in  slag . 29,820 

Heat  in  pig  iron •  • 32,500 

Heat  in  cooling  water 14,922 

Lost  by  radiation  and  conduction  (diff.) .  37,791 

344,835 

(6)  Compare  the  heat  items  which  are  substantially  different 
for  the  furnace  run  by  moist  and  dried  blast. 

Moist  Blast.  Dried  Blast. 

Combustion  of  C  to  CO 131,220  92,950 

Heat  in  waste  gases 43,836  23,799 

Decomposing  moisture  in  blast 14,511  3,225 

Loss  by  radiation  and  conduction 51,180  37,791 

It  may  be  noticed,  that  using  moist  blast  too  much  carbon 
was  burnt  to  CO  at  the  tuyeres;  the  chief  item  of  economy 
with  dried  blast  is  the  ability  to  get  along  with  less.  The 
smaller  total  amount  of  gases,  particularly  nitrogen,  accounts 
for  the  lower  temperature  of  the  waste  gases,  with  dried  blast. 
The  direct  saving  by  reason  of  decomposition  of  the  moisture 
is  the  smallest  item  of  economy.  The  reduced  losses  by  radia- 
tion and  conduction  are  mainly  because  of  the  faster  rate  of 
running,  these  losses  being  nearly  constant  per  day.  The  ratio 
of  these  losses  is  found  to  be  0.74,  whereas,  the  inverse  ratio 
of  the  pig  iron  productions  per  day  was  0.79. 

(7)  Calculate  the  carbon  burnt  at  the  tuyeres,  the  propor- 
tion of  the  carbon  available  thus  consumed.     Compare  these 
items  with  those  of  the  furnace  on  moist  blast. 

Moist  Blast.     Dried  Blast. 

Carbon  burnt  at  tuyeres 75.3  58 . 05 

Total  fixed  carbon  charged 84 . 3  67 . 8 

Proportion  burnt  at  tuyeres 89 . 3  85 . 6 

Fixed  carbon  really  available 79 . 4  62.9 

Proportion  burnt  at  tuyeres 94 . 8  92 , 3 


RATIONALE  OF  HOT-BLAST  AND  DRY-BLAST 


307 


(In  some  charcoal  furnaces  of  low  heat  requirement,  i.e., 
with  pure  ores  and  fuel,  as  little  as  37  parts  of  carbon  is  burned 
at  the  tuyeres  per  100  of  pig  iron  produced,  which  represents, 
moreover,  only  70  to  75  per  cent,  of  the  available  fixed  carbon 
in  these  furnaces.) 


Charges. 

Pig  Iron. 

Slag. 

Gases. 

CO 

fc 

Fe2O3     135.7 
H2O          17  8 

Fe     95.0 



0        40.7 
H2O   17.8 

£ 

0 

SiO2          17.8 
A12O3         6  3 

Si        1.0 

SiO2      15.7 
A12O3     6  .  3 

0          1.1 

SiO2           9  2 

SiO2       2  2 

3 

CaO          21  1 

CaO      21  .  1 

% 

MgO           2  1 

MeO       2  1 

E 

CO2          19.0 

CO2    19.0 

C               67.8 

C         4.0 

C        63.8 

fc 

SiO2           4  2 

SiO2       4  .  2 

v» 

CaO            4  2 

CaO        4  2 

Z 

H2O            0.8 

H2O     0.8 

U5 

O2             76  .  5 

O        76.5 

n 

N2           255  0 

N      255.0 

f 

H2O            1  0 

f  H         0.1 

5 

\  O          0.9 

Totals    631.5 

100.0 

55.8 

475.7 

(8)  The  proportion  of  the  whole  heat  requirement  available 
in  the  region  of  the  tuyeres. 

Answer. — 53  per  cent. 

(9)  The  proportion  of  the  iron  in  the  furnace  which  is  re- 
duced by  solid  carbon  from  FeO. 

Answer. — 23.8  per  cent. 

(10)  The  theoretical  maximum  of  temperature  at  the  tuyeres. 
Answer.— 1965°  C. 


308  METALLURGICAL  CALCULATIONS. 

HOT  BLAST. 

For  several  centuries  blast  furnaces  were  run  by  charcoal  as 
fuel  and  with  cold  blast.  How  great  the  variations  in  tem- 
perature of  the  cold  blast  may  be,  can  be  inferred  from  the 
experience  of  a  furnace  manager  in  the  Urals,  Russia,  who 
noted  temperatures  of  the  air  nearly  40°  C.  in  the  summer 
and  —  60°  C.  in  the  winter.  Assuming  an  average  temperature 
of  0°  C.  for  unheated  blast,  burning  charcoal  to  CO,  the  theo- 
retical maximum  of  temperature  before  the  tuyeres  can  be 
calculated  as  follows: 

Heat  generated  by  combustion  2430  Calories 

Heat  in  hot  carbon  being  burnt        0.5t—    120 

Volume  of  CO  and  N2  formed  5.3795  cubic  meters 

2310  +  0.5t 

Temperature  ==  5.3944  (o.303  +  Q.Q00027t) 

Whence  t  =  1678° 

This  does  not  mean  that  the  pig  iron  and  slag  will  be  car- 
ried to  this  temperature,  any  more  than  if  a  locomotive  could 
run  alone  90  miles  an  hour  it  could  therefore  pull  a  train  of 
cars  that  fast.  The  hot  gases,  CO  and  N2,  are  at  the  start  at 
this  temperature,  and  as  they  ascend  and  come  in  contact  with 
the  descending  iron  and  slag,  these  are  raised  to  temperatures 
approximating  towards,  but  always  lower  than,  the  above. 
In  fact,  the  temperature  -to  which  the  iron  and  slag  is  raised 
depends  on  the  relative  quantities  of  iron  and  slag  to  fuel  burnt, 
and  the  speed  with  which  the  furnace  is  worked. 

If  the  blast  is  heated,  its  sensible  heat  is  simply  added  to  the 
numerator  of  the  above  expression.  We  can  easily  find  out 
how  much  sensible  heat  the  4.4685  cubic  meters  of  air  brings 
in,  at  any  temperature  desired  [Q  =  4.4685  (0.303t  +  0.000027t2)], 
and  then  solve  the  quadratic  anew.  We  thus  find 

Temp,  of  Blast.  Heat  in  Blast.     Theoretical  Temp. 

0°  C.  -  Calories  1678°  C. 

100°  "  137         "  1762°  " 

200°  "  276         "  1845°  " 

300°  "  417         "  1929°  " 

400°  "  561         "  2012°  " 

500°  "  707         "  2096°  " 

600°  "  856         "  2180°  " 


RATIONALE  OF  HOT-BLAST  AND  DRY-BLAST          309 

Temp,  of  Blast.  Heat  in  Blast.  Theoretical  Temp. 

700°  C.  1007  Calories.         2265°  C. 

800°  "  1160       "  2350°  " 

900° "  1316       "  2435°  " 

1000°  "  1475       "  2520°  " 

The  heating  of  the  blast  thus  raises  the  maximum  tempera- 
ture available  some  85°  C.  for  each  100°  C.  to  which  the  blast 
is  heated.  It  not  only  increases  the  temperature  available,  but 
also  the  number  of  heat  units,  thus  increasing  both  the  quan- 
tity and  the  intensity  of  the  heating  in  the  tuyere  region.  Of 
these  two  items  of  increase,  that  of  the  intensity  factor  is  the 
most  important,  since  it  regulates  the  rapidity  of  transfer  of 
heat  to  the  charge  and  the  efficiency  and  speed  of  the  smelting 
action  of  the  furnace. 

DRIED  BLAST. 

Each  kilogram  of  water  vapor  decomposed  absorbs  29040  •*• 
9  =  3227  Calories,  which  would  not  be  needed  if  the  0.67  kilo, 
of  carbon  thus  employed  was  oxidized  by  air  instead  of  by 
moisture.  Per  kilogram  of  carbon  oxidized  by  water  vapor, 
there  is  absorbed  58,080 -=-12  =  4,840  Calories,  while  this  kilo- 
gram of  carbon  can  only  furnish  2430  Calories  in  becoming 
CO,  leaving  a  net  absorption  of  2410  Calories  per  kilogram  of 
carbon  thus  burned,  against  which,  however,  can  be  credited 
the  heat  in  the  kilo,  of  carbon  burned  and  the  sensible  heat  in 
the  water  vapor  itself;  the  former  is  0.5t—  120,  and  the  latter 
can  be  calculated  from  the  volume  of  the  water  vapor,  1.8519 
cubic  meters.  We  thus  have,  per  kilogram  of  carbon  thus 
oxidized : 

Temperature 

of  Water  Vapor.  Heat  in  Moisture.  Heat  in  Products. 

100°  66  Calories  0.5t-  2460    Calories 

200°  137         "  0.5t-2393 

300°  214  0.51-2316 

400°  296         "  0.5t-2234 

500°  384         "  0.51-2146 

600°  478         "  0.5t-2042 

700°  577         "  0.51-1953 

800°  682         "  0.51-1848 

900°  792         "  0.51-1738 

1000°  907         "  0.51-1623 


310  METALLURGICAL  CALCULATIONS. 

Since  the  heat  in  the  carbon  burnt  (0.5t—  120)  can  never 
equal  numerically  1623,  it  is  seen  that  under  no  circumstances 
can  the  water  vapor  do  anything  but  diminish  the  quantity  of 
heat  available  at  the  tuyeres,  while  the  products  of  its  decom- 
position CO  and  H2  increase  the  volume  of  the  products  and 
so  diminish  still  further  the  intensity  of  temperature  attainable. 

The  best  way  to  determine  the  amount  of  moisture  in  the'  air 
is  to  draw  it  through  a  tube  containing  concentrated  sulphuric 
acid,  measure  the  quantity  of  dry  air  drawn  through,  and 
weigh  the  amount  of  moisture  caught  by  the  tube.  This  gives 
the  weight  of  moisture  accompanying  unit  volume  of  dry  air 
(not  weight  of  moisture  in  unit  volume  of  moist  air).  De- 
terminations by  the  wet  and  dry  bulb  thermometers,  the  whirled 
psychrometer,  humidity  gauges,  etc.,  are  none  of  them  so  re- 
liable as  the  above  described  method,  which  is  direct,  simple 
and  accurate.  The  results  may  be  obtained  in  grains  per 
cubic  foot  of  dry  air  or  milligrams  per  liter.  The  best  way  to 
express  them,  for  further  use,  is  in  ounces  avoirdupois  per 
cubic  foot,  or  kilograms  per  cubic  meter.  The  first  is  obtained 
"by  dividing  the  number  of  grains  by  437.5,  the  second  by  divid- 
ing the  milligrams  per  liter  by  1000.  The  numbers  thus  ob- 
tained are  practically  identical  in  the  two  systems. 

The  theoretical  temperatures  attainable  with  moist  blast  of 
varying  degrees  of  moisture  and  heated  to  various  tempera- 
tures are  obtained  by  applying  the  previously  explained  prin- 
ciples. We  have  already  calculated  the  temperature  obtained 
by  burning  carbon  with  dry  air  of  various  temperatures.  We 
have  also  calculated  a  table  of  the  deficit  of  heat  available  pro- 
duced by  the  entrance  of  1.5  kilos,  of  water  vapor  (which  would 
oxidize  1  kilo,  of  carbon  and  produce  1.8519  cubic  meters  of 
CO  and  H2).  We  are,  therefore,  prepared  to  calculate  a  table 
of  the  maximum  temperature  attainable  using  blast  of  any 
degree  of  humidity  heated  to  any  practical  temperature.  Before 
giving  the  table  we  will  run  through  the  details  of  one  calcula- 
tion, to  make  clear  the  method  employed. 

Illustration. — What  is  the  theoretical  maximum  tempera- 
ture using  air  which  carries  normally  10  grams  of  moisture  per 
cubic  meter  of  dry  air,  reduced  to  standard  conditions  (i.e.  ,10 
grams  per  1.293  kilograms  of  dry  air),  and  heated  to  500°  C.? 

It  takes  3.5275  cubic  meters  of  dry  air  at  standard  condi- 


RATIONALE  OF  HOT-BLAST  AND  DRY-BLAST          311 

tions  to  burn  1  kilogram  of  carbon.  If  dry,  and  at  500°,  there 
is  2430  Calories  generated  by  combustion,  707  Calories  in  the 
dry  air  used,  0.5t—  120  Calories  in  the  hot  carbon,  and  the 
total  heat  thus  at  hand  raises  the  5.3944  cubic  meters  of  pro- 
ducts to  the  temperature  of  2096°  C.,  as  is  determined  by  solving 
the  equation 

2430  +  707+  (0.5t-  120) 


5.3944  (0.303 +  0.000027t) 


=  2096° 


As  modified  by  the  moisture,  the  4.4685  cubic  meters  of  dry 
air  would  be  accompanied  by  44.685  grams  of  moisture,  or 
0.044685  kg.,  which  would  oxidize  two-thirds  of  its  weight  of 
carbon,  or  0.02979  kg.  of  carbon,  which  would  contribute  to  the 
heat  available  0.02979  (0.5t-2146)  Calories,  making  a  con- 
tribution of  0.015t—  64  Calories  to  the  numerator  of  above 
equation.  The  products  of  combustion  will  be  increased  by  H2 
and  CO  equal  in  volume  to  twice  the  volume  of  the  moisture, 
or  2 X  0.044685  H- 0.81  =  0.1102  cubic  meters,  the  mean  specific 
heat  of  which  goes  into  the  denominator.  We  then  have 

=  3017  +  0.5t  +  0.015t-64 

"  (5.3944  +  0.1102)  (0.303 +  0.000027t)  " 

Another  method  of  solution,  not  using  the  previously  calcu- 
lated tables  but  based  entirely  on  first  principles,  is  to  base  the 
whole  calculation  on  the  use  of  1  cubic  meter  of  dry  air,  with 
its  accompanying  moisture,  as  follows: 

Oxygen  present  in  1  m3  dry  air  =  1,293X3/13       =  0.2984  kg. 
Oxygen  present  in  the  moisture  =  0.010X8/9         =  0.0089    " 

Total  =  0.3073    " 

Carbon  consumed  =  0.3073X0.75  =  0.2305   " 

Volume  of  moisture  =  0.010  H- 0.81  =  0.0123  m3 

Volume  of  oxygen  in  dry  air  =  0.2078   ' 

Volume  of  products  from  dry  air  =  1.0000  +  0.2078  =  1.2078    " 
Volume  of  products  from  moisture  =  2X0.0123      =  0.0246    " 

Total  volume  of  products  =  1.2324    " 

Heat  of  combustion  of  carbon  =  0.2305X2430       =        560  Cal. 
Heat  in  carbon  at  t°  =  0.2305X  (0.5t-  120)  =  0.1152t-  28     " 


312  METALLURGICAL  CALCULATIONS. 

Heat  in  dry  air  at  500°  =  IX  [0.303  +  0.000027 

(500)]  500  158  Cal. 

Heat  in  moisture  at  500°  =  0.0123  [0.34  +  0.00015 

(500)]  3 

Heat  absorbed  in  decomposing  moisture  =  0.0123 

X2614  =  -32  " 

Whence  results  the  equation 

0.1152t  +  661 


0.2324  (0.303  +  0.000027t) 


By  applying  either  of  the  two  methods  of  calculation  ex- 
plained, the  temperatures  in  the  following  table  are  obtained: 

THEORETICAL  TEMPERATURES  BEFORE  THE  TUYERES 

j  Grams  per  cubic  meter  of   dry  air  reduced  to  0°  C. 
MOISTURE,  j  Grains  per  cubic  foot  of  dry  air  re(iuce(}  to  0°  C. 

Grams  5.00     10.00     15.00      20.00      25.00      30.00      35.00      40.00 


Grains.  2.19      4.38      6.57         8.75       10.94       13.13       15.32       17.50 


Blast. 

40° 

1678° 

1647° 

1615° 

1573° 

1536° 

1507° 

1471° 

1443° 

100° 

1723 

1692 

1666 

1627 

1587 

1548 

1526 

1496 

200° 

1807 

1776 

1751 

1712 

1673 

1636 

1612 

1584 

300° 

1892 

1861 

1837 

1800 

1760 

1725 

1700 

1673 

400° 

1976 

1945 

1921 

1885 

1846 

1813 

1786 

1760 

500° 

2061 

2030 

2007 

1970 

1933 

1902 

1874 

1848 

600° 

2146 

2115 

2093 

2055 

2020 

1991 

1962 

1936 

700° 

2232 

2201 

2178 

2144 

2108 

2080 

2050 

2025 

800° 

2318 

2287 

2264 

2227 

2195 

2169 

2138 

2114 

900° 

2404 

2373 

2351 

2313 

2282 

2257 

2226 

2203 

1000° 

2490 

2459 

2437 

2399 

2369 

2345 

2314 

2292 

The  calculations  show  that  the  temperature  before  the  tuyeres 
may  vary  as  much  as  235°  C.  =  423°  F.,  from  change  in  the 
moisture  of  the  air,  from  dryness  to  warm  air  saturated  with 
moisture. 


CHAPTER  VI. 
PRODUCTION,  HEATING  AND  DRYING  OF  AIR  BLAST. 

This  is  a  subject  intimately  related  to  the  running  of  iron 
blast  furnaces,  and  incidentally  related  more  or  less  to  all 
classes  of  metallurgical  operations.  The  principles  involved 
are  those  of  physics  —  mechanical  and  thermal  —  and  when  once 
thoroughly  understood  can  be  used  for  the  most  various  classes 
of  metallurgical  problems. 

PRODUCTION  OF  BLAST. 

There  are  two  different  principles  upon  which  air  is  com- 
pressed, represented  by  the  fan  and  the  piston  machines.  The 
first  is  constant  in  its  operation,  the  second  intermittent,  the 
first  draws  in  and  expels  a  continuous  current  of  air,  the  second 
draws  in  a  given  quantity  of  air,  compresses  it  and  expels  it. 
Measuring  the  work  done  by  the  difference  in  static  conditions 
of  the  compressed  and  of  the  uncompressed  air,  the  work  actu- 
ally done  is  a  fixed  and  calculable  quantity,  independent  of 
what  type  of  machine  performs  it.  During  the  compression, 
heat  is  generated,  and  the  mechanical  work  done  includes  the 
mechanical  equivalent  of  the  heat  thus  generated.  This  is 
allowed  for  in  the  well-known  formula  for  adiabatic  com- 
pression : 


in  which 

f     =  the  ratio  of  the  specific  heat  of  air  at  constant  pres- 

sure to  that  at  constant  volume  =  1.408. 
V0  =  the  volume  of  the  uncompressed  air. 
p0   =  the  tension  of  the  uncompressed  air. 
pi   =  the  tension  of  the  compressed  air. 

If  we  use  the  value  of  f  =  1.408,  the  coefficient  —  —  r  be- 

r—  1 
comes  3.45,    and  -  -  =0.29.     If  we  then  use  the  other  di- 

r 

313 


314  METALLURGICAL  CALCULATIONS. 

mensions  in  feet  and  pounds,  the  resulting  work  done  will  be 
in  foot-pounds;  if  we  use  meters  and  kilograms,  in  kilogram- 
meters.  The  element  of  time  does  not  enter  into  this  equation, 
the  work  actually  done  is  the  same  for  compressing  a  given 
amount  of  air,  and  is  independent  of  the  time.  If  I  lift  a  kilo- 
gram 1  meter  high,  the  same  amount  of  work  is  done  whether 
I  lift  it  in  1  minute  or  in  1  second,  the  rate  of  doing  work  would 
be,  however,  sixty  times  as  great  in  the  last  instance  as  in  the 
first,  but  the  actual  amount  of  work  accomplished  is  the  same 
in  each  case.  If,  therefore,  we  use  in  the  formula  the  volume 
of  air  entering  the  compressor  per  minute,  the  result  will  give 
the  work  done  per  minute;  if  we  use  the  volume  per  second 
we  get  the  work  done  per  second.  If  we  wish  to  transpose 
the  work  done  into  horse-power,  we  bear  in  mind  that  a  horse- 
power is  understood  in  English  units  as  33,000  foot-pounds  of 
work  done  per  minute,  and  in  the  metric  system  as  75  kilogram- 
meters  of  work  done  per  second. 

In  applying  the  formula  we  may  further  notice  that  —  ex- 

Po 

presses  the  ratio  of  compression;  that  is,  by  how  many  times 
the  tension  of  the  original  air  is  increased.  If  air  at  one  at- 
mosphere tension  (such  as  the  air  we  usually  have  to  breathe) 
is  compressed  to  two  atmospheres  tension,  the  ratio  of  com- 
pression is  2;  if  air  entered  a  machine  at  two  atmospheres  ten- 
sion and  was  therein  compressed  to  four  atmospheres,  the 
ratio  of  compression  would  be  likewise  2,  and  the  work  done 
(for  a  given  quantity  of  air)  the  same  as  before.  The  effective 
pressure,  as  shown  on  a  gauge,  is  the  difference  between  these 
two  tensions:  it  is  not  pj.  It  is  highly  important  that  this 
point  be  held  clearly  in  mind.  The  tension  of  the  uncompressed 
air  is  its  barometric  pressure,  as  measured  against  a  vacuum; 
the  tension  of  the  compressed  air  is  likewise  its  pressure  as 
measured  against  a  vacuum;  the  effective  pressure  of  the  com- 
pressed air  is  the  difference  between  these  two  tensions;  the 
tension  of  the  compressed  air  is,  therefore,  (by  transposition) 
equal  to  the  barometric  pressure  of  the  uncompressed  air  plus 
the  effective  pressure  of  the  compressed  air,  i.e.,  plus  the  gauge 
pressure.  Never  use  the  effective  pressure  of  the  compressed 
air  as  pl5  but  always  add  to  it  the  barometric  pressure  of  the 
uncompressed  air  p0  for  the  correct  value  of  pt. 


PRODUCING    HEATING  AND  DRYING  BLAST.          315 

Regarding  the  volume  of  uncompressed  air,  V0  is  its  volume 
measured  at  the  pressure  P0,  i.e.,  the  actual  volume  of  uncom- 
pressed air  at  its  actual  tension.  If  more  convenient,  however, 
we  may  use  the  volume  of  this  air  measured  at  standard  baro- 
metric pressure,  if  we  multiply  it  by  p0,  the  standard  pres- 
sure. Thus,  if  we  know  the  volume  of  uncompressed  air  meas- 
ured at  one  atmosphere  standard  tension,  we  may  use  for  the 
p0  immediately  following  it  the  standard  pressure  10.334  (kilos, 
per  square  meter),  or  2,120  (pounds  per  square  foot).  By  the 
law  of  the  reciprocal  nature  of  volume  and  pressure,  the  pro- 
duct thus  obtained  is  the  same  as  would  be  found  by  multi- 
plying the  volume  of  the  same  air  at  any  other  pressure  by 
that  pressure  (V0  p0  =  Vj  px  =  V2  p2,  etc.).  Whatever  pres- 
sure we  use  in  getting  the  product  V0  p0,  however,  we  must 
use  only  the  actual  tension  of  the  uncompressed  air  as  the 
denominator  in  getting  the  ratio  of  compression  (p!-v-p0). 

Problem  59. 

A  blast  furnace  requires  2,615  cubic  meters  of  air,  measured 
at  —  5°  C.,  for  every  metric  ton  of  pig  iron  made.  Assuming 
outside  barometric  pressure  735  millimeters  of  mercury  column, 
the  efficient  pressure  of  the  compressed  blast  to  be  1.0  kilo- 
gram per  square  centimeter  (as  measured  in  a  blast  regulating 
reservoir) ,  a  mechanical  efficiency  of  the  blowing  engine  of 
90  per  cent.,  of  delivery  of  blast  82.3  per  cent.,  and  output  447 
tons  per  day: 

Required : 

(1)  The  ratio  of  compression  of  the  blast. 

(2)  The  piston  displacement  per  ton  of  pig  iron  made. 

(3)  The  work  of  the  blowing  cylinders,  per  ton  of  iron  made. 

(4)  The  gross  or  indicated  horse-power  of  the  steam  cylinders. 
Solution : 

(1)  The   effective   pressure   being    1.0   kilogram   per   square 

1  0000 

centimeter,  represents  0004^  760  =  735  millimeters  of  mer- 
cury. But  the  uncompressed  air  is  at  735  mm.  barometric 
tension,  therefore,  the  total  tension  of  the  compressed  air,  plf  is 
1470  mm.  of  mercury,  and  the 

Ratio  of  compression  =  =2.0  ^1) 

7oo 


316  METALLURGICAL  CALCULATIONS. 

(2)  Piston  displacement  per  metric  ton  of  pig  iron: 

g-g23  =       3165  cubic  meters.  (2) 

=  111,560  cubic  feet. 

(3)  The  volume  displaced  must  be  the  basis  of  calculating 
the  power  required,  since  subsequent  losses  of  air  (100—  82.3 
=  17.7  per  cent.)  do  not  affect  the  work  done  in  the  blowing 
cylinder.     This  volume  is  3,165  cubic  meters,  representing  that 
volume  of  uncompressed  air  at  —  5°  C.  and  735  mm.  pressure. 
The  initial  temperature  of  the  air  does  not  affect  the  work  to 
be  done,  and  we  can  either  use  its  volume  at  735  mm.  tension 
or  correct  this  to  760  mm.  tension,  just  as  we  chose.     If  we 
consider  the  volume  at  735  mm.  tension,  then  the  product  of 
volume  and  pressure  is 

Vp  =  3,165  X  (lO,334  x  ~\  =  31,755,000 
If  we  first  change  the  volume  to  standard  pressure,  we  have 

(70K  \ 
3,165  X  ^-j  X  10,344  =  31,755,000 

The  work  done  in   compression,   in   the  blowing   cylinder  is, 
therefore, 

W  =  3.45X3,073X10,334X[2°-29-  1] 

=  109,554,750  X[l. 2225  -  1] 

=  24,326,800  kilogrammeters.  (3) 

(4)  The  steam  in  the  steam  cylinders  does  more  mechanical 
work  than  would  be  represented  by  the  compression  of  air  in 
the  blowing  cylinders.     In  this  case,  we  assume  90  per  cent, 
mechanical  efficiency,   or   10  per  cent,   mechanical  loss.     The 
gross  work  of  the  steam  is,  therefore, 

24,326,800^-0.90  =  27,029,800  kilogrammeters, 
or,  in  horse-power, 

27,029,900X447 


1440X4500 


=  1,865  horse-power. 


This  amounts  to  4.17  h.p.  per  metric  ton  of  pig  iron  made 
per  day. 


PRODUCING,  HEATING  AND  DRYING  BLAST.          317 

MEASUREMENT  OF  PRESSURE. 

In  making  calculations  of  the  work  done  in  furnishing  blast, 
as  in  the  preceding  problem,  it  is  important  to  note  how  the 
pressure  of  the  compressed  air  is  measured.  The  real  pressure 
is  measured  correctly  by  a  pressure  gauge  only  where  the  air 
is  comparatively  at  rest,  as  on  a  pressure-equalizing  reservoir. 
If  the  pipe  communicating  with  a  gauge  is  connected  with  a 
blast  main,  in  which  the  velocity  of  the  air  is  considerable,  the 
pressure  recorded  will  vary  greatly  with  the  direction  of  the 
end  of  this  tube  relative  to  the  direction  of  the  air  current. 
The  total  pressure  of  the  air  in  motion  is  the  pressure  which  it 
exerts  against  the  sides  of  the  main,  plus  the  pressure  which 
has  been  used  in  giving  it  velocity.  If  the  pressure  gauge 
tube  is  cut  off  at  right  angles  to  the  flow  of  air  in  the  main, 
the  pressure  recorded  is  even  less  than  the  actual  static  pres- 
sure of  the  air  against  the  sides  of  the  main  (because  of  suction 
effect),  and  does  not  include  at  all  the  pressure  represented 
by  the  velocity.  The  only  way  to  measure  correctly  in  a 
main  the  total  pressure  which  has  been  impressed  upon  the 
blast  is  to  bend  the  end  of  the  pressure  tube  until  it  is  parallel 
with  the  axis  of  the  main  and  faces  the  current  of  air.  The 
gauge  then  records  the  static  pressure,  plus  the  velocity  head, 
and  gives  the  proper  pressure  to  use  in  calculating  the  work  done 
by  the  blowing  engine.  However,  it  is  still  preferable  to  put 
the  gauge  on  a  blast  reservoir,  if  there  is  one,  where  the  air 
is  nearly  at  rest  and  velocity  head  approximates  zero — for 
the  reason  that  the  velocity  of  air  passing  through  a  tube  is 
greatest  in  the  center  and  least  against  the  walls,  and  it  is  diffi- 
cult to  place  the  pressure  tt:be  so  as  to  measure  the  mean  ve- 
locity. An  "approximation  to  the  mean  value  is  found  by 
placing  the  end  of  the  pressure  tube  not  in  the  center  but  at 
one-third  of  the  radius  of  the  main  from  the  center. 

INDICATOR  CARDS. 

If  the  blowing  engine  cylinder  can  be  tested  by  pressure 
indicator  cards,  then  a  different  method  of  obtaining  the  work 
of  producing  the  blast  is  furnished.  The  integration  of  the 
diagram  gives  the  mean  pressure  on  the  piston  during  the 
stroke.  Expressing  this  in  kilograms  per  square  meter  or 
pounds  per  square  foot,  and  multiplying  by  the  area  of  the 


318  METALLURGICAL  CALCULATIONS. 

piston  and  the  length  of  the  cylinder,  we  get  the  work  done 
per  stroke;  again  multiplying  by  the  number  of  strokes  per 
minute  we  have  the  work  done  per  minute,  from  which  is  readily 
obtained  the  horse-power.  These  operations  are  usually  com- 
bined in  the  formula: 

Work  =  PXLXAXN 

If  we  observe  that  LxAXN  is  the  piston  displacement  per 
minute,  the  formula  becomes: 

Work  =  P  X  Piston  displacement  per  minute, 

in  which,  it  must  not  be  forgotten,  P  represents  mean  pres- 
sure on  the  piston  during  the  stroke,  and  is  not  to  be  con- 
founded with  gauge  pressure  of  the  compressed  air.  Such  a 
formula  is,  therefore,  totally  inapplicable  to  finding  the  work 
done  by  a  fan  or  rotary  blower,  in  which  only  the  final  pressure 
of  the  compressed  air  is  known.  The  mistake  of  so  applying 
it  is  often  made.  When  only  the  final  pressure  of  the  compressed 
air  is  known,  the  formula  for  adiabatic  compression  is  the  only 
correct  one  to  use. 

It  may  not  be  useless  to  call  attention  to  the  fact  that  when 
using  the  formula  for  adiabatic  compression,  the  raising  of  the 
ratio  of  compression  to  the  0.29  power  has  to  be  done  by  logar- 
ithms : 


If  a  table  of  logarithms  is  not  at  hand,  a  satisfactory  ap- 
proximation may  be  made  by  taking  the  cube  root  of  the  ratio, 
since 


HEATING  OF  THE  BLAST. 

Air  blast  is  commonly  heated  either  continuously,  by  direct 
transmission  of  heat  through  metal  or  earthenware  pipes,  or 
discontinuously  by  the  heating  up  of  fire-brick  surfaces,  which 
are  subsequently  cooled  by  the  blast.  It  is  not  within  the  prov- 
ince of  these  calculations  to  enter  into  a  description  of  the 
various  types  of  such  hot-blast  stoves,  but  to  indicate  the 
principles  upon  which  the  metallurgist  can  base  calculations  as 
to  the  efficiency  of  such  stoves,  and  thus  be  prepared  to  find 


PRODUCING,  HEATING  AND  DRYING  BLAST.          319 

out  which  do  the  best  work,  and  wherein  lie  the  principal  ad- 
vantages or  disadvantages  of  each  type. 

The  efficiency  of  a  hot-blast  stove  is  measured  by  the  ratio 
of  the  heat  imparted  to  the  blast  to  that  contained  in  the  air 
and  fuel  used  and  generated  by  combustion.  It  is  a  furnace 
whose  useful  effect  is  the  heat  imparted  to  the  air,  and  all  other 
items  of  heat  distribution  are  more  or  less  necessary  losses. 
The  problem  is  simplest  when  the  stove  is  a  recuperator  (con- 
tinuous type),  using  solid  fuel.  Then  the  item  of  heat  genera- 
tion is  perfectly  definite,  since  the  amount  of  fuel  used  in  a 
given  time  can  be  definitely  determined.  When  the  hot-blast 
stove  receives  gaseous  fuel,  however,  from  a  blast  furnace,  the 
amount  of  gas  received  by  the  stoves  is  usually  a  very  uncertain 
quantity,  since  only  part  of  all  the  gas  produced  by  the  furnace 
is  used  by  the  stoves,  and  the  question  of  what  fraction  is  thus 
used  is  difficult  to  determine.  In  such  cases,  the  usual  method 
of  comparing  the  sizes  of  the  pipes  leading  gas  to  the  stoves 
and  those  carrying  gas  to  boilers,  etc.,  affords  but  a  very  un- 
certain determination,  since  the  draft  and  consequent  velocity 
of  the  gas  may  be  very  different  in  the  two  sets  of  pipes.  In 
such  cases,  knowing  the  total  volume  of  gas  produced  by  the 
furnace,  not  only  the  relative  sizes  of  the  hot-blast  stove  pipes 
should  be  noted,  but  also  the  relative  velocities  of  the  gas 
currents  in  the  two  sets  of  pipes.  Another  method  is  to  de- 
termine the  quantity  of  chimney  gas  passing  away  from  the 
stoves  into  the  chimney  flues,  by  measuring  the  size  of  the 
chimney  flue,  temperature  of  the  gases  and  average  velocity; 
given  in  addition  the  chemical  analyses  of  the  chimney  gas  and 
of  the  furnace  gas,  the  quantity  of  the  latter  being  used  can  be 
calculated,  using  the  carbon  in  the  two  gases  as  the  basis  of 
calculation. 

Problem  60. 

Statement:  Air  for  drying  peat  in  a  kiln  is  heated  from 
0°  C.  to  150°  C.  by  an  iron  pipe  stove,  the  latter  consuming 
dried  peat,  whose  ultimate  analysis  is: 

Carbon 49.70  per  cent. 

Hydrogen ».     5.33       " 

Oxgyen 30.76       " 

Nitrogen 1.01       " 

Ash..  .  13.23       " 


320  METALLURGICAL  CALCULATIONS. 

The  calorific  power  of  this  peat  is  4249,  and  by  the  com- 
bustion of  92.5  kilograms,  5122  cubic  meters  of  air  is  heated 
to  the  required  temperature.  The  products  of  combustion  pass 
away  from  the  stove  at  200°  C.,  and  contain  (analyzed  dry) 
14.8  per  cent,  of  CO2,  and  no  CO  or  CH4  or  other  gas  containing 
carbon. 

Required: 

(1)  The  heat  generated  in  the  stove. 

(2)  The  heat  in  the  hot  air,  and  the  net  efficiency  of  the 
stove. 

(3)  The  volume  and  composition  of  the  products  of  com- 
bustion in  the  stove. 

(4)  The  heat  carried  out  in  the  chimney  of  the  stove. 

(5)  The  heat  lost  by  radiation  and  combustion. 

(6)  The  excess  of  air  used  in  burning  the  peat. 

* 

(1)  Heat  generated  in  the  stove: 

4249X92.5  =  393,033  Calories       (1) 

(2)  Heat  imparted  to  the  air: 

[0.303  +  0.000027  (150)]  X  150  X 

5122  =  235,907 

Net  efficiency  =        60.0  per  cent.     (2) 

(3)  Volume  of  products  of  combustion: 

Weight  of  carbon  in  fuel  burnt—  92.5 

X 0.4970  =       45.97  kilos. 

Volume  of  CO2  thus  produced-  45.97 

-7-0.54  =      85.13  cub.  meters 

Volume  of  (dry)  gas  produced—  85.13 

-i-0.148  =    575.20     " 

Volume  of  N2  and   O2  in   products  — 

575.2-85.13  =    490.07     " 

Volume  of  water  vapor  in  products  — 

92.5  X  0.0533  X  9 -=- 0.81  =      54.78     " 

Percentage  composition  of  products  of  combustion: 

Moist.  Dried. 

CO2 13.5  14.8 

N2andO2 77.8  85.2 

H20..  .  8.7 


PRODUCING,  HEATING  AND  DRYING  BLAST.          321 

To  separate  the  nitrogen  and  oxygen  would  necessitate  con- 
siderable calculation,  and  is  not  required  just  here,  because 
the  two  gases  have  the  same  heat  capacity  per  cubic  meter, 
and  therefore  can  thermally  be  reckoned  together: 

(4)  Heat  in  the  chimney  gases  at     200°: 

CO2  85.13  m3X0.4140  =    35.24 

N2  and  O2  490.07  m3X  0.3084  =  151.14 
H2O  54.78  m3X  0.3700  =    20.27 


206.65X200  =  41,330  Cal. 
Proportion  thus  lost  =  10.5  P.  C.     (4) 
(5)  Loss  by  radiation  and  conduction 

100  -  (60.0  + 10.5)  =  29.5  P.  C.     (5) 

(6)  The  separation  of  N2  and  O2  in  the  products  is  not  easy. 
It  is  best  based  on  the  consideration  that  part  of  these  con- 
sists of  the  nitrogen  of  the  air  which  was  necessary  for  com- 
bustion (plus  the  nitrogen  of  the  fuel)  and  of  the  excess  air. 
The  first  can  be  calculated,  and  thus  the  latter  obtained  by 
difference,  and  then  the  percentage  of  air  in  excess  determined. 
Oxygen  necessary  for  combustion: 

00 

C  to  CO2  =  45.97X7=  =  122.59  kilos, 

iz 

Hto  H20  =  4.93X8  =    39.44     " 

162.03     " 
Oxygen  present  in  peat  92.5X0.3076  =    28.45     " 


Oxygen  to  be  supplied  =  133.58 

Nitrogen  accompanying  =  445.27 


Air  necessary  =  578.85 

=  447.7    m3 

Nitrogen  from  coal,  92.5X0.0101  =      0.93  kilos. 

Total  nitrogen  from  coal  and  necessary  air       =  446.2 
Volume  =  446.2-r-l.26  =354.1    m3 

Total  N2  and  O2  in  products  =  490.1      " 

Therefore,  excess  air  =  136.0     " 

Percentage  of  excess  air  ^-^  =  0.303  =    30.3  P.  C. 

(6) 


322  METALLURGICAL  CALCULATIONS. 

Problem  61. 

A  bla$t  furnace  has  three  hot-blast  stoves,  two  of  which  are 
always  on  gas  and  one  on  blast.  Per  long  ton  of  pig  iron  pro- 
duced there  issues  from  the  furnace  (reduced  to  standard  tem- 
perature and  pressure) : 

Nitrogen 81,763  cubic  feet 

Carbon  monoxide 25,383 

Carbon  di-oxide 20,409      " 

Water  vapor 8,230 

For  the  same  quantity  of  pig  iron  made,  92,330  cubic  feet 
(standard  conditions)  of  air  is  heated  in  the  stoves  from— 5° 
C.  to  465°  C.  The  hot  gases  reach  the  stoves  at  175°  C.,  are 
there  burned  by  10  per  cent,  excess  of  air  at  0°,  and  the  chimney 
gases  resulting  (assume  perfect  combustion)  pass  out  of  the 
stoves  at  120°  C. 

Required: 

(1)  The  net  efficiency  of  the  stoves,  assuming  they  receive 
25,  30,  35,  40,  45  or  50  per  cent,  of  the  gas  produced  by  the 
furnace. 

(2)  Assuming   that    each    stove   radiates   and   loses   to   the 
ground  one-third  as  much  heat  as  the  blast  furnace  itself  (the 
heat  balance  sheet  of  the  furnace  showed  846,518  pound  Cal- 
ories thus  lost  per  long  ton  of  pig  iron  produced),  what  pro- 
portion of  the  whole  gas  goes  to  the  hot-blast  stoves,  and  what 
is  the  net  efficiency  of  the  stoves? 

Solution : 

(1)  We  will  first  calculate  the  efficiency  of  the  stoves,  as- 
suming that  they  received  all  the  gas  produced,  from  which 
datum  can  then  be  obtained  the  efficiency,  supposing  any  as- 
sumed fraction  of  the  gas  goes  to  the  stoves. 
Heat  in  the  blast  ( -5°  to  465°) : 

92,330  X [0.303  +  0.000027  (465  -5)]  X  [465  -  (  -5)] 

=  92,330    XO. 31542X470  =  13,687,683  ounce  cal. 

=        855,480  pound  Cal. 
Sensible  heat  in  the  hot  gases  (0°  to  175°) : 

N2  and  CO     107,146X0.3077  32,969 

CO2  20,409X0.4085  8,337 

H2O  8,230X0.3763  3,097 

Heat  capacity  per  1°  44,403  ounce  cal. 

2,775  pound  CaL 
Heat  capacity  per  175..  2,775X175°  =      .485,625      " 


PRODUCING,  HEATING  AND  DRYING  BLAST.          323 

Heat  generated  by  combustion: 

CO  to  CO2  25,383X3062  =  77,722,746  ounce  cal. 

4,857,672  pound  Cal. 

Total  heat  available  =     5,343,297       " 

If  the  gas  were  all  used  in  the  stoves  the  efficiency  of  the 
latter  would  be  only 

-      855,480 


5,343,297 


=  0.160  =  16.0  per  cent. 


With  smaller  proportions  of  the  whole  gas  used  the  efficiency 
of  the  stoves  calculates  out  as  follows: 

Using  50  per  cent,  of  the  gas Efficiency  32.0  per  cent. 

"45         "  "  "          36.0       " 

"40         "  "  "          40.0       " 

"35         "  "  45.7       " 

"30         "  "  "          53.3 

"25         "  "  .......         "          64.0 

"      16         "  "  "        100.0       " 

(1) 

All  that  the  above  analysis  tells  us  is,  that  certainly  over 
16  per  cent,  of  the  gas  produced  by  the  furnace  iriust  be  used 
by  the  stoves ;  but  if  we  can  deduce  any  probable  value  for  the 
percentage  of  the  gas  actually  used,  such  as  by  measuring  the 
several  gas  mains  and  the  velocities  of  the  gas  in  each,  we  can 
then  reckon  out  the  value  of  the  efficiency  of  the  stoves,  with 
about  the  same  degree  of  probability.  Blast  furnaces  use 
from  33  to  60  per  cent,  of  the  gas  they  produce  in  the  stoves. 
If  we  assume  50  per  cent.,  in  this  case,  the  efficiency  of  the 
stoves  would  be  32  per  cent. 

(2)  There  is  another  way  of  solving  the  problem,  which  is 
to  either  measure,  calculate  or  assume  the  heat  lost  by  radia- 
tion and  conduction  from  the  stoves;  adding  to  this  the  heat 
going  out  in  the  products  of  combustion,  and  the  net  heat  in 
hot  blast,  the  sum  is  the  total  heat  which  has  been  brought 
into  and  generated  within  the  stoves.  But,  all  the  gas  would 
bring  in  and  generate  in  the  stoves  5,343,297  Calories;  we  can, 
therefore,  find  easily  what  proportion  of  all  the  gas  is  being 
used  in  the  stoves.  In  requirement  (2)  we  are  told  to  as- 
sume that  the  three  stoves  lost  by  radiation  and  conduction 


324  METALLURGICAL  CALCULATIONS. 

846,518  pound  Calories  per  long  ton  of  pig  iron  made,  an  amount 
equal  to  that  similarly  lost  by  the  furnace  itself.  The  heat 
in  the  chimney  gases,  at  120°,  can  be  thus  calculated: 

Oxygen  required  ( —  CO  j  =     12,692  cubic  feet 

Air  required  =    61,017  " 

Excess  of  air  used  =      6,102  " 

Nitrogen  of  required  air  =     48,325  "         " 

Nitrogen  already  in  gas  =     81,763  "         " 

Nitrogen  in  these  two  items  =  130,088  " 

Chimney  products: 

CO3  =  45,792  " 

H2O  =  8,230  " 

N2  =  130,088  " 

Excess  air  =  6,102  " 

Heat  in  chimney  gases : 

CO2  45,792X0.3964  =  18,152  ounce  cal. 

H2O  8,230X0.3580  =  2,946     " 

N2  and  air    136,190X0.3064  =  41,729     " 

Heat  capacity  per  1°  =     62,827     " 

=       3,927  pound  Cal. 
Heat  capacity  per  120°  =  471,122     " 

If  all  the  furnace  gas  were  burnt  in  the  stoves,  471,122  pound 
Calories  would  go  up  the  stove  chimneys.  But,  since  only  a 
part  of  the  gas  is  so  used,  only  a  fraction  of  this  amount  of 
heat  is  lost  to  the  chimneys.  If  we  call  X  the  proportion  of 
the  gases  used,  then  471,122  X  will  represent  the  chimney  loss, 
and  the  total  heat  requirements  of  the  stoves  will  be: 

Heat  in  air  blast  =      855,480  Ib.  Cal. 

Heat  in  chimney  gases          =      471,122  X    " 
Radiation  and  conduction    =      846,518 

Total  =  1,701,998+471,122  Xlb.  Cal. 
The  proportion  of  the  total  gases  needed  to  supply  this,  X.  is 

1,701,998  +  471, 122  X 
5,343,297 

whence 

X  =  0.349  =  34.9  per  cent.  (2) 


PRODUCING,  HEATING  AND  DRYING  BLAST.  325 

and  the  net  efficiency  of  the  stoves  is 
855,480  855,480 


5,343,297  X  0 . 349      1,864,812 


=  45.9  per  cent.   .  (2) 


Arithmetically,  the  finding  of  X  can  be  simplified,  perhaps, 
by  considering  that  the  chimney  loss  represents  in  any  case 
471,122-^-5,343,297  =  0.088  =  8.8  per  cent,  of  the  total  heat 
received  by  the  stoves,  leaving  91.2  per  cent,  to  be  applied  to 
heating  the  blast  and  for  radiation  and  conduction  loss.  The 
last  two  items,  however,  must  amount  to  1,701,998  Calories, 
and,  therefore,  the  total  heat  requirement  of  the  stoves  is 


1'866'200  P°und 
requrng 

iggO=  0.349  .  34.9  per  cent.  (2) 

of  all  the  gas  produced  by  the  furnace. 

In  the  above  solution  the  only  uncertain  factor  in  the  calcu- 
lation is  the  radiation  and  conduction  loss  from  the  stoves,  and 
this  uncertainty  does  not  largely  affect  the  reliability  of  the 
result  obtained,  allowing  that  we  have  assumed  an  approxi- 
mately correct  value  for  this  loss.  All  uncertainty  could  be 
dispelled,  however,  were  the  radiation  and  conduction  losses 
measured  directly.  This  could  be  accomplished  by  finding 
experimentally  the  temperature  of  the  outside  shells  of  the 
stoves,  and  calculating  the  external  losses  of  heat  by  the  laws 
of  radiation  and  conduction;  but  in  order  to  do  this  satisfac- 
torily, it  would  be  necessary  to  divide  the  shells  of  the  stoves 
into  zones,  and  determine  the  temperature  and  calculate  the 
losses  from  each  zone  separately  —  a  rather  long  operation,  but 
one  worth  doing. 

DRYING  AIR  BLAST. 

The  advantage  of  dried  blast  has  been  already  discussed  at 
length  in  these  calculations.  The  advantage  is  due  primarily 
to  the  higher  temperature  obtained  when  the  moisture  has  been 
removed. 

The  means  adopted  commercially  for  drying  the  air  have 


326  METALLURGICAL  CALCULATIONS. 

been  those  of  refrigerating  the  uncompressed  air,  before  its 
entrance  into  the  blowing  cylinders.  This  has  the  great  ad- 
vantage of  furnishing  the  cylinders  with  cold  air,  and,  there- 
fore, of  greatly  increasing  their  blowing  capacity,  since  the 
weight  or  quantity  of  air  blown  is  proportional  to  the  absolute 
temperature  of  the  air  entering  the  cylinders. 

Illustration.  Outside  air  being  at  30°  C.,  what  increase  in 
the  amount  of  air  furnished  by  blowing  engines  will  result  if 
the  temperature  of  the  air  is  artificially  reduced  to  —5°  C.  ? 
How  much  slower  can  the  engines  be  driven,  in  the  second 
case,  in  order  to  furnish  the  same  weight  of  air  as  before? 

The  two  temperatures  are  respectively  273  +  30  and  273—5 
absolute,  that  is,  303  and  268.  The  engines,  if  run  at  uniform 
speed,  would  furnish  303-7-268  =1.13  times  as  much  air  in 
the  second  case,  or  13  per  cent.  more.  If  the  engines  were 
slowed  down  to  268-^-303  =  0.884  of  their  former  speed  they 
would  furnish  the  same  amount  of  air;  that  is,  they  could  be 
run  11.6  per  cent,  slower.  In  fact,  they  could  be  run  more 
than  11.6  per  cent,  slower  in  the  second  case,  and  yet  supply 
the  same  quantity  of  air,  because  at  the  slower  running  the 
delivery  efficiency  is  somewhat  higher. 

The  disadvantage  of  cooling  the  uncompressed  blast  is  that 
it  must  be  cooled  much  more,  to  eliminate  from  it  a  given  per- 
centage of  its  moisture,  than  if  it  were  first  compressed,  and 
to  obtain  nearly  dry  air  the  moisture  must  be  practically  frozen 
out. 

Illustration:  Air  at  30°  C.  and  85  per  cent,  humidity  is  to 
be  cooled  until  95  per  cent,  of  its  moisture  is  eliminated,  with- 
out compression;  to  what  temperature  must  it  be  cooled? 

From  the  tables  of  aqueous  tension,  we  find  that  the  maxi- 
mum tension  which  aqueous  vapor  can  exert  at  30°  C.  is  31.5 

31  5 

millimeters,    which    means   practically   that     ^  '      of   a    cubic 

•  oU 


pr 

meter  of  moisture  accompanies         '     of  a  cubic  meter  of  air 

proper.     If  the  humidity  is  85  per  cent.,  then  this  same  quantity 
of  air  is  accompanied  by 

31  5 

=  0.0352  cubic  meters 


PRODUCING,  HEATING  AND  DRYING  BLAST.          327 
of  moisture,  or,  per  cubic  meter  of  air  measured  dry — 


0. 0352  -*-         -    =  0.0368  cubic  meters. 


If  95  per  cent,  of  this  is'  removed  by  cooling,  then  the  mois- 
ture left,  per  cubic  meter  of  air  measured  dry,  is 

0.0368X0.05  =  0.00184  cubic  meters. 

And  the  relative  tensions  of  air  and  moisture,  to  attain  this 
dryness,  must  have  been  reduced  to 

1.0000:0.00184. 

Since  the  sum  of  these  tensions  is  always  760  mm.  in  the  un- 
compressed air,  the  actual  tensions  of  air  and  moisture  will  be 

758.6:1.4, 

that  is,  the  temperature  must  be  reduced  until  the  moisture 
present  can  exert  only  1.4  mm.  pressure.  This,  on  examining 
the  tables  of  tension  of  aqueous  vapor  is  found  to  be  less  than 
0°  C.,  in  fact -15°  C. 

Problem  62. 

Air  at  30°  C.,  carrying  85  per  cent,  of  its  possible  amount  of 
moisture,  is  cooled  to  0°  C.,  and  the  moisture  condensed  to 
liquid  water  at  that  temperature.  Barometer  760  mm. 

Required : 

(1)  The  percentage  of  the  moisture  condensed. 

(2)  The  amount  of  heat  to  be  abstracted  from  each  cubic 
meter  of  original  moist  air  (refrigerating  effect). 

(3)  The  percentage  of  moisture  which  would  be  condensed 
if  the  temperature  were  reduced  to  —5°  C. 

(4)  The  total  heat  to  be  abstracted,  per  cubic  meter  of  orig- 
inal air,  in  the  latter  case. 

Solution : 

(1)  From  the  preceding  illustration  we  can  take  the  figures 
that  in  the  moist  air  taken,  each  cubic  meter  of  air  proper  is 
accompanied  by  0.0368  cubic  meter  of  moisture.  In  the  cooled 


328  METALLURGICAL  CALCULATIONS. 

air  at  0°,  the  relative  volumes  of  air  proper  and  residual  moisture 
will  be  as  their  relative  tensions,  viz.: 

755.4:4.6 
or 

1:0.0061. 

The  moisture  accompanying  the  given  quantity  of  air  is,  there- 
fore, reduced  from  0.0368  to  0.0061,  showing  a  condensation 
of  0.0307,  equal  to 


(2)  The  air  at  30°  contains,  as  before  figured  out,  air  proper 
and  moisture  in  the  proportions 

1:0.0368. 

or,  in  1  cubic  meter  of  moist  air,  in  the  proportions 
0.9645:0.0355. 

We  have,  therefore,  to  calculate  the  heat  to  be  extracted  from 
0.9645  cubic  meter  of  air  proper,  falling  30°  to  0°,  and  from 
0.0355  cubic  meter  of  water  vapor,  falling  30°  to  0°  and  83.4 
per  cent,  of  the  latter  condensing  to  liquid  at  0°.  The  figures 
are,  remembering  that  the  volumes  heretofore  handled  are 
at  30°: 

Air  proper: 

070 
0  .  9645  X        X  0.3038  X  30  =     7  .  920  Calories 


Moisture,  if  all  uncondensed: 

oyq 

0.0355X^X0.3445X30  =    0.332 

oUo 

Heat  of  condensation: 


0.0355X0.834XX0.81X606.5  =  13.105       " 

oUo  _ 

Total  =  21.  357       "      (2) 

(3)  If  the  temperature  were  reduced  to  —5°  C.,  the  tension 
of  the  residual  moisture  would  be  3.4  mm.  of  mercury,  and 


PRODUCING,  HEATING  AND  DRYING  BLAST.          329 

that  of  the  air  proper  756.6  mm.,  their  relative  volumes  would 
be  in  the  same  ratio,  viz.: 

756.6:3.4 
or 

1:0.0045. 

showing  that  out  of  0.0368  cubic  meter  of  moisture  originally 
accompanying  1  cubic  meter  of  air  proper,  0.0323  had  been 
condensed,  or 


0   0393 

0.878  =  87.8  per  cent.  (3) 


(4)  The  condensation  is  seen  to  be  87.8—83.4,  or  4.4  per 
cent,  more,  if  the  air  is  cooled  to  —  5°  C.  A  considerably  larger 
amount  of  refrigeration,  however,  is  required  in  this  case, 
because  the  moisture  would  all  be  frozen.  The  1  cubic  meter 
of  moist  air  at  30°,  containing,  as  before  calculated,  0.9645 
cubic  meter  of  air  proper  and  0.0355  cubic  meter  of  moisture, 
would  have  87.8  per  cent,  of  the  latter  condensed  to  ice,  or 
0.03117  cubic  meter,  and,  therefore,  0.00433  cubic  meter  re- 
maining uncondensed. 

Air  proper,  30°  to  -5°  C.: 

070 

0  .  9645  X  57^  X  0.3037  X  35  =    9  .  237  Calories 

oUo 

Moisture,  if  all  uncondensed,  30°  to  —5°  C.: 

070 

0.0355X5^X0.3438X35  =    0.385       " 

oUo 

Condensation  of  0.03177  m3  to  liq.  at  -  5°  C.: 

070 
0.03177X5^X0.81X605  =  14.025       " 

oUo 

Freezing  of  same,  at  —  5°  C.  : 


27Q 
0.03177X^X0.81X80 

OUO 


(4) 


The  conclusion  is,  that  an  increased  condensation  of  4.4  per 
cent,  has  been  obtained  by  an  increase  in  the  refrigerating 
requirement  of  25.502—21.357  =  4.145  Calories,  or  nearly 


330  METALLURGICAL  CALCULATIONS. 

20  per  cent.  Another  way  of  stating  the  comparison,  is  that 
when  not  freezing  the  moisture  condensed,  about  4  per  cent, 
of  the  moisture  was  condensed  from  each  cubic  meter  of  air 
for  one  Calorie  of  refrigeration,  whereas,  the  removal  of  ad- 
ditional moisture  by  cooling  below  zero  requires  nearly  one 
Calorie  refrigeration  for  each  additional  per  cent,  of  moisture 
eliminated. 

A  practical  conclusion  is,  that  the  expense  of  refrigeration 
might  easily  be  justified  down  to  0°  C.,  and  yet  be  unjustified 
by  the  results  when  cooling  below  0°. 

Mr.  James  Gayley  has,  I  believe,  patented  the  scheme  of 
refrigerating  the  air  in  two  stages;  first,  nearly  to  zero,  re- 
moving the  moisture  thus  condensed  as  liquid,  and  then  cooling 
the  nearly  dry  air  further,  eliminating  more  moisture  as  ice. 
In  this  manner,  the  amount  of  refrigeration  required  is  reduced 
by  the  latent  heat  of  solidification  of  the  larger  part  of  the 
moisture,  and  cooling  below  zero  becomes  profitable.  The 
saving  in  the  above  example  would  be  80  Calories  per  kilogram 
on  all  the  moisture  condensed  at  0°,  viz.: 

0.0216X80  =  1.728  Calories, 

cutting  down  the  refrigeration  required  from  25.502  to  23.774 
Calories,  that  is,  enabling  4.4  per  cent,  additional  drying  to  be 
produced  for  2.4  Calories  additional  refrigeration.  This  scheme 
of  Mr.  Gayley  is  founded  on  sound  scientific  as  well  as  prac- 
tical considerations. 

The  idea  of  cooling  the  compressed  blast  by  moderately 
cool  water,  recently  proposed,  is  also  a  very  practical  idea 
and  founded  on  sound  scientific  principles.  At  a  given  tem- 
perature moisture  cannot  exert  more  than  a  maximum  vapor 
tension.  It  follows  that  if  we  start  with  air  saturated  with 
moisture,  at  a  given  temperature,  and  compress  it  to  double  its 
initial  tension,  keeping  its  temperature  constant,  about  half  of 
its  moisture  must  condense  out.  If,  at  the  same  time,  it  is 
artificially  cooled,  then  more  than  half  of  its  moisture  will 
condense  as  liquid. 

If  we  start  with  air  not  saturated  with  moisture,  the  com- 
pression, temperature  being  constant,  will  increase  the  tension 
of  the  moisture  present  until  the  air  becomes  saturated,  after 
which  increased  pressure  will  cause  condensation. 


PRODUCING,  HEATING  AND  DRYING  BLAST.          331 

Illustration:  Air  at  30°  C.,  85  per  cent,  saturated  with  mois- 
ture, is  compressed.  At  what  effective  pressure  does  it  become 
saturated  with  moisture,  temperature  remaining  constant? 

The  moisture  present  is  exerting  85  per  cent,  of  its  maximum 
vapor  tension  at  this  temperature.  Therefore,  the  tension 
must  be  increased  in  the  ratio  of  85  to  100,  to  make  the  air 
saturated,  viz.:  in  the  ratio  1  to  1.177.  The  effective  pressure 
necessary  to  be  applied  is,  therefore,  1.177  —  1.000  =  0.177 
atmospheres  (2.6  pounds  per  square  inch). 

Illustration:  If  air  at  30°  C.,  85  per  cent,  saturated  with 
moisture,  is  compressed  to  one  atmosphere  effective  pressure 
and  its  temperature  kept  'constant,  what  proportion  of  its 
moisture  will  condense? 

Before  compression,  the  tension  of  the  moisture  being  31. 5 X 
0.85  =  26.8  mm.,  the  relative  volumes  of  air  proper  and  mois- 
ture are  as 

733.2  :  26.8 
or,  as 

1  :    0.0367 

After  compression,  the  tension  on  the  mixture  being  two 
atmospheres  (1520  mm.),  and  the  tension  of  the  uncondensed 
moisture  being  it  maximum,  or  31.5  mm.,  the  relative  volumes 
of  air  and  uncondensed  moisture  will  be  as 

1488.5  :  31.5, 
or,  as 

1  :    0.0212. 

The  proportion  of  the  original  moisture  remaining  uncondensed 
is  therefore, 

-  °-578  - 57-8  per  cent- 

and  the  amount  condensed  out  =  42.2  per  cent. 

Problem  63. 

Air  at  30°  C.,  and  85  per  cent,  saturated  with  moisture  is 
compressed  to  one  atmosphere  effective  pressure  (760  mm.  of 
mercury),  and  simultaneously  cooled  by  river  water  to  10° C. 
Barometer  730  mm. 

Required : 

(1)  The  percentage  of  the  original  moisture,  condensed. 


332  METALLURGICAL  CALCULATIONS. 

(2)  The  weight  of  moisture  remaining  in  the  air,  expressed 
in  grams  per  cubic  meter  of  dry  air  at  standard  conditions 
(i.e.,  per  1.293  kilograms  of  air  proper). 

Solution :     (1) 

Tension  of  uncompressed  moist  air  =    730 . 0  mm. 

Tension  of  moisture  present  31.5X0.85  =      26.8     " 

Tension  of  air  proper,  uncompressed  =     703.2     " 

Relative  volumes  of  air  proper  and  moisture      =  1  :  0.0381 
Tension  of  compressed  moist  air  730  +  760          =  1490.0     " 
Tension  of  uncondensed  moisture  (maximum 

tension  at  10°)  .  =        9.1     " 

Tension  of  air  proper,  when  compressed  =  1480.9     " 

Relative  volumes  of  air  proper  and  uncondensed 

moisture  =  1  :  0.0061 

Proportion  of  moisture  condensed: 

0.0381-0.0061  .n 

0.0381          =  °'84  =  a4° Per  Cent'  (1) 

(2)  The  relative  volumes  of  air  proper  and  uncondensed 
moisture  are,  as  found  above, 

1  : 0.0061, 

And  the  relative  specific  gravities  of  air  proper  and  moisture 
are  as  the  standard  weights  of  1  cubic  meter  of  each,  viz.:  as 

1.293  : 0.81. 

It  follows,  therefore,  that  1.293  kilos,  of  air  proper  (1  cubic 
meter  at  standard  conditions)  must  be  accompanied  by, 

0.0061X0.81  =  0.0049  kg.  of  moisture, 
or       *  =  4.9  grams.  (2) 

The  original  moist  air  contained,  similarly  considered, 

0.0381X0.81  =    0.0309kg. 
—  30.9  grams. 


CHAPTER  VII. 
THE  BESSEMER  PROCESS. 

The  outlines  of  this  famous  process  are  known  to  every 
educated  person;  the  mechanical  and  most  of  the  chemical 
details  are  familiar  to  most  technologists;  the  exact  way  to  run 
the  converter  is  the  source  of  income  to  hundreds  of  superin- 
tendents of  works,  and  yet  the  quantitative  side  of  the  chemical 
and  physical  operations  involved  is  mastered  by  very  few. 

To  state  the  case  briefly,  melted  pig  iron  is  put  into  the  con- 
verter, numerous  air  jets  are  blown  through,  the  impurities  of 
the  iron — carbon,  silicon,  manganese  and,  in  a  special  case, 
phosphorus — oxidize  relatively  faster  than  the  iron,  and  the 
final  product  is  usually  nearly  pure  iron.  This  is  recarburized 
to  steel  by  spiegel-eisen.  During  the  blow  very  little  free 
oxygen  escapes  from  the  converter,  and  the  gases  produced  are 
principally  nitrogen,  carbon  monoxide  and  some  carbon  di- 
oxide, while  some  hydrogen  may  come  from  the  decomposition 
of  the  moisture  of  the  air.  The  silicon,  manganese,  phosphorus 
and  iron  form  silica,  manganous  oxide  (MnO)  principally,  fer- 
rous oxide  (FeO)  principally,  phosphorus  pentoxide,  P2O5, 
which  go  into  the  slag,  while  a  little  Fe203,  Mn3O4  and  SiO2 
escape  as  fume. 

The  applications  of  calculations  to  this  process  are  numerous 
and  important.  They  include  such  subjects  as  the  amount  of 
air  theoretically  required  per  ton  of  iron,  the  dimensions  and 
power  of  the  blowing  engines,  the  weight  of  slag  produced,  the 
balance  sheet  of  materials,  the  balance  sheet  of  heat  evolved 
and  distributed,  the  radiation  losses,  the  discussion  of  the 
efficiency  of  the  various  impurities  as  heating  agents  in  the 
process. 

AIR  REQUIRED. 

Basing  our  calculations  on  the  analysis  of  the  pig  iron  used, 
and  assuming  it  to  be  blown  to  pure  iron,  we  must  next  as- 
sume the  probable  loss  of  iron  itself  in  the  blow.  This  varies 

33? 


334  METALLURGICAL  CALCULATIONS. 

considerably,  from  1  to  10  per  cent,  on  the  pig  iron  used  in 
ordinary  practice,  but  as  much  as  20  to  25  per  cent,  in  some 
carelessly  run  "  Baby  "  Besserners  in  steel-casting  foundries. 
The  silicon  all  oxidizes  to  SiO2;  iron  mostly  to  FeO,  and  a 
small  part,  say  not  over  one-tenth,  to  Fe2O3;  manganese  mostly 
to  MnO,  a  small  part,  up  to  one-fourth,  may  form  Mn2O3; 
phosphorus  forms  only  P2O5;  carbon  burns  mostly  to  CO,  but 
from  one-fifth,  at  times  nearly  one-half,  burns  to  CO2.  When 
all  the  calculated  oxygen  has  been  found,  the*blast  to  contain 
it  can  be  calculated,  if  it  is  assumed  that  no  free  oxygen  es- 
capes from  the  converter;  at  times,  however,  up  to  one-third 
of  all  the  oxygen  blown  in  may  thus  escape,  but  this  is  very 
exceptional,  ordinarily  less  than  one-fifth  thus  escapes,  and 
often  none  at  all. 

Problem  64. 

Pig  iron  containing  3.10  per  cent,  carbon,  0.98  silicon,  0.40 
manganese,  0.101  phosphorus  and  0.06  sulphur  is  blown  in  an 
acid-lined  converter,  to  metal  practically  free  from  carbon, 
silicon  and  manganese,  but  no  sulphur  or  phosphorus  is  elimi- 
nated. To  get  the  minimum  and  the  maximum  amounts  of 
air  which  could  be  needed,  make  the  following  assumptions: 

Case  1.         Case  2. 
Per  cent,  of  iron  lost  by  oxidation. . . ._. ... ...     1  15 

Proportion  of  iron  oxidized  to  Fe2O8. . ...none        one-tenih 

Proportion  of  Mn  oxidized  to  Mn203 .  ...none        one-fifth 

Proportion  of  C  oxidized  to  CO2.  /. ..one-fifth  one-half 

Proportion  of  O2  escaping  in  the  gases '.none        one-third 

Requirement:  (1)  Find  the  weight  of  dry  air  needed  per 
metric  ton  of  pig  iron  blown,  in  each  case,  and  its  volume  at 
0°  C.  Express  the  results  also  in  pounds  and  cubic  feet  per 
ton  of  2,000  pounds. 

Solution : 

Case  1. 

Oxygen  needed  per  1,000  kg.  of  pig  iron: 

00 

C  to  CO2 6.2  kg.X ||  -     16.53  kg. 

CtoCO..  ....24.8   "   X—  =    33.07    " 


THE  BESSEMER  PROCESS.  335 

00 

Si  to  SiO2  .......................   9.8   "    X^i  =     11.20  kg. 

Zo 

MntoMnO  ......................   4.0    "    X^|  =       1.16   " 

oo 

FetoFeO  .......................  10.0   "   X       =      2.86   " 


64.82    " 
N2  accompanying  =  216.07    " 


Air  needed =  280.89      " 

Volume  at  0°  C. .......  i =   217.2     m3 

Volume  needed  per  1000  oz.  Av =  217.2  ft3 

Volume  needed  per  2000  Ibs.  Av =  6,950  ft3 

Case  2. 
Oxygen  needed" per  1000  kg.  of  pig  iron: 

qo 

C  to  CO2 15.5  kg.X j^  =    41.33  kg. 

CtoCO.... 15.5   "   X^  =    20.67   " 

iZ 

Si  to  SiO2..  .   9.8   "   x||=     11.20   " 

Zo 

1  fi 

MntoMnO 3.2    "    X^  =       0.93    " 

55 

Mn  to  Mn2Q3 . .  0.8  "  X  ^=      0.34    " 

FetoFeO..  .135.0    MX~—    38.57    " 

OD 

FetoFe203 15.0   "   X^=      6.43   " 

O2  unused  (one-half  sum  of  above) =    59.73    " 


179.20    ' 
N2  accompanying  =  597.33    " 


Air  used =  776.53    ' 

Volume  at  0°  C =       600  m3 

Volume  used  per  1000  oz.  Av =        600  ft3 

Volume  used  per  2000  Ibs.  Av =  19,200  ft3 


336  METALLURGICAL  CALCULATIONS. 

For  temperatures  of  the  air  above  0°  C.,  a  corresponding 
increase  in  the  volume  used  would  be  found.  Since  this  is  net 
air  received  by  the  converter  an  allowance  for  loss  of  10  to 
25  per  cent,  (in  exceptional  cases  50  per  cent.)  would  be  needed 
to  get  the  piston  displacement  of  the  blowing  engines.  The 
above  figures  are  the  maximum  and  minimum  for  this  quality 
of  pig  iron  only,  blown  in  an  acid-lined  converter;  other  quali- 
ties of  pig  iron  might  require  a  little  more  or  less,  and  if  blown 
in  a  basic-lined  converter  considerably  more,  to  oxidize  the 
phosphorus.  The  detailed  calculations  can  be  made  in  each 
specific  case. 

AIR  RECEIVED. 

The  converse  of  the  preceding  proposition  is  to  take  an 
actual  case,  in  a  Bessemer  converter,  and  to  calculate  how 
much  air  is  being  received.  This  will  serve  as  a  check  on  the 
blowing  engines,  since  the  volume  received,  divided  by  the 
piston  displacement,  gives  the  volume  efficiency  of  the  blowing 
plant.  To  make  the  calculation  we  need  to  know  the  weight 
and  analysis  of  the  pig  iron  and  the  analysis  of  the  blown  metal, 
in  order  to  find  the  weights  of  impurities  oxidized,  also  the 
average  composition  of  the  escaping  gases,  to  find  the  propor- 
tion of  carbon  burning  to  CO2  and  of  unused  oxygen;  also  the 
composition  of  the  slag,  to  get  therefrom  the  weight  of  iron 
oxidized  and  the  weight  of  slag,  if  practicable,  but  this  can 
sometimes  be  calculated;  also  the  weight  and  composition  of 
the  fume,  if  it  is  considerable.  The  temperature  of  the  air 
entering  the  blowing  cylinders  its  hygrometric  condition  and 
the  barometric  pressure  should  also  be  noted. 

Problem  65. 

At  the  South  Chicago  works  of  the  Illinois  Steel  Co.  (see 
paper  by  Howe,  in  Trans.  American  Institute  of  Mining  Engi- 
neers, XIX.  [1890-91],  p.  1127),  the  charge  weighed  22,500 
pounds,  and  contained  2.98  per  cent,  carbon,  0.94  silicon,  0.43 
manganese,  0.10  phosphorus,  and  0.06  sulphur.  After  blowing 
9  minutes  10  seconds  the  bath  contained  0.04  per  cent,  carbon, 
0.02  silicon,  0.01  manganese,  0.11  phosphorus  and  0.06  sulphur. 
The  slag  formed  contained  63.56  per  cent,  silica,  3.01  alumina, 
21.39  FeO,  2.63  Fe203,  8.88  MnO,  0.90  CaO  and  0.36  MgO,  of 


THE  BESSEMER  PROCESS.  337 

which  the  APO3,  Cal  and  MgO  and  part  of  the  SiO2  come  from 
the  lining.  The  gases,  analyzed  dry,  show  an  average  com- 
position during  the  blow  of 

CO2 5.20  per  cent. 

CO 19.91 

H2 1.39 

N2 ...73.50 

and  were  free  from  fume. 

The  piston  displacement  during  the  blow  was  190,406  cubic 
feet,  air  in  engine  room  36°  C.,  humidity  50  per  cent.,  barometer 
756  millimeters. 

Requirements:  (1)  The  weight  of  oxygen  acting  on  the  bath 
during  the  blow,  and  the  theoretical  volume  of  air  at  standard 
conditions  to  which  this  would  correspond,  per  2,000  pounds 
of  metal  blown. 

(2)  The  volume  of  moist  air  at  the  conditions  of  the  engine 
room,  received  by  the  converter  during  the  blow,  and  the  vol- 
ume efficiency  of  the  blowing  machine  and  connections. 

(3)  The  weight  of  slag  produced  and  the  loss  in  weight  of 
the  lining  by  corrosion  during  the  blow. 

Solution:  (1)  The  percentages  of  impurities  left  in  the  bath 
are  so  small  that  we  can  take  them  as  equivalent  to  the  same 
percentages  reckoned  on  the  original  weight  of  the  bath  If 
they  had  been  larger  it  would  be  necessary  to  assume  an  ap- 
proximate loss  of  iron  during  the  blow,  find  the  final  weight 
of  the  bath  and  reckon,  the  percentages  on  this  revised  weight. 

Making  this  assumption,  we  know  at  once  the  weights  of 
carbon,  silicon  and  manganese  oxidized,  but  we  do  not  know 
the  weight  of  iron  lost.  That  follows,  however,  from  a  con- 
sideration of  the  slag,  for  the  manganese  and  iron  in  the  slag 
are  derived  only  from  the  metallic  bath,  and  the  analysis  of  the 
slag  practically  gives  us  the  relation  between  the  weights  of 
manganese  and  iron  in  it;  since  we  know  the  weight  of  the 
former  oxidized,  the  weight  of  iron  lost  can  be  calculated. 
Thus  the  slag  contains : 

MnO        8.88  per  cent  =    6.88  per  cent.  Mn 

FeO        21.39       "          =  16.64        "          Fe  as  FeO 

Fe203       2.63       "         =     1.84        "         Fe  as  Fe203 


338  METALLURGICAL  CALCULATIONS. 

The  loss  of  manganese  being  0.42  per  cent.  =  94.5  pounds,  the 
loss  of  iron  is : 


Ifi  fi4 
94.5 X-^   '-  =  228.6  pounds  Fe  as  FeO 

O.oo 


94.5  X  =  25-3  pounds  Fe  as  Fe2O3 

b.oo 

The  remaining  item  still  undetermined  is  the  weight  of  carbon 
oxidizing  to  CO2  and  to  CO.  The  gas  analysis  shows  5.20  per 
cent.  CO2  to  19.91  per  cent.  CO,  and  ^since  equal  volumes  of 
each  of  these  gases  contain  equal  weights  of  carbon,  it  follows 

5  20 

that       '         of  the  total  carbon  is  present  in  the  gas  as  CO2, 
Zo.iL 

and  the  rest  as  CO.     Since  the  total  carbon  oxidized  is  22,500  X 
0.0294  =  661.5  pounds,  we  have 


661.5  X  =  137.0  pounds  C  burning  to  CO2 


=  524.5       "  "  CO 

Weight  of  oxygen  absorbed  by  the  bath  : 

C  to  CO2  ...........  137.0  pounds  X   8/3  =  365.3  pounds 

CtoCO  ............  524.5       "       X   4/3  =699.3 

SitoSiO2  ..........  207.0       "       X32/28  =236.6 

MntoMnO  .........   94.5       "       X  16/55  =    27.5       " 

FetoFeO  ..........  228.6       "       X  16/56  =    65.3       " 

FetoFe203  .......   25.3       "       X48/112  =     10.8 


1404.8      " 
N2  corresponding =  4682.7 

Dry  air  corresponding =  6087.5 

Volume  at  0°  C =  75,483  ft8 

Weight  of  O2  per  2000  pounds =  124.9  lbs.(l) 

Volume  of  air  per  2000  pounds =  6,710  ft3(l) 

(This  result  is  for  comparison  with  data  of  Problem  64). 


THE  BESSEMER  PROCESS.  339 

(2)  The  nitrogen  in  the  gases  can  be  obtained  from  its  vol- 
ume relation  to  the  carbon,  and  from  this  we  can  calculate  the 
real  volume  of  blast  used. 

Weight  of  carbon  in  1  cubic  foot  of  gases: 

(0.0520  + 0.1991)  X  0.54  =  0.1356  oz.  Av. 

Volume  of  gases  produced  at  standard  conditions: 

661.5X16     _7gQ,7  f,3 
0.1356         78'°57  J 

Volume  of  nitrogen  at  standard  conditions: 

78,057X0.7350  =  57,372  ft3 
Weight  =  57,372X1.26      =  72,289  oz.  Av. 

=     4,518  Ibs. 

The  question  now  is,  how  much  nitrogen  is  contained  in  each 
cubic  foot  of  air  in  the  engine  room.  Knowing  that,  we  are 
prepared  to  calculate  the  volume  of  this  actually  received  by 
the  converter:,  » 

Barometric  pressure  =        756  m.m. 

Tension  of  moisture  (44x0.5)  =         22     " 


Tension  of  air  present  =        734 

Tension  of  nitrogen  present  (734X0.792)         =       580 
Weight  of  nitrogen  in  1  cubic  foot  : 


97*3 

=  0.8495  OZ.Av 


Volume  of  air  actually  received : 

4,518X16 
0.8495 

Volume  efficiency  of  machinery: 
85,095 


85,095  ft.3  (2) 


190,406 


=  0.447  -    44.7  p.  c.  (2) 


(3)  The  slag  contains  8.88  per  cent,  of  MnO,  equal  to  6.88 
per  cent,  of  Mn,  as  already  calculated.     But  94.5  pounds  of 


340  METALLURGICAL  CALCULATIONS. 

manganese  is  oxidized,  therefore,  the  weight  of  slag  produced 

is: 

94.5^-0.0688  =  1374  pounds.  (3) 

Weight  of  Si02in  slag  (1374  X  0.6356)  =  873  pounds 

SiO2  from  the  Si  of  bath  (207.0  +  236.6)  =444 


SiO2  corroded  from  the  lining  =  429 

CaO,  A12O3  and  MgO  (1374X0.0427)  =     59 


Loss  in  weight  of  lining  =  488       "       (3) 

BLAST  PRESSURE. 

It  is  necessary  to  use  sufficient  blast  pressure  to  overcome 
the  static  pressure  of  the  metallic  bath,  plus  that  of  the  slag 
formed,  also  the  back  pressure  in  the  converter,  to  give  the 
necessary  velocity  to  the  air  in  the  tuyeres  and  to  overcome 
friction  in  the  same.  When  the  tuyeres  are  near  to  the  sur- 
face of  the  bath,  pressures  of  1  or  2  pounds  will  run  the  small 
converter,  but  the  orcjmary  converter  with  bottom  tuyeres  re- 
quires from  15  to  30  pounds  pressure  per  square  inch  (1.054  to 
2.108  kg.  per  square  c.m.).  We  will  consider  the  latter,  the 
more  frequent  and  the  more  complex  case  to  discuss. 

The  metal  lies  12  to  24  inches  (30  to  60  c.m.)  deep  in  the 
converter.  Since  its  specific  gravity  melted  is  about  6.88  (Rob- 
erts and  Wrightson),  the  ferro-static  pressure  which  it  exerts 
is  practically  0.25  pounds  per  square  inch  for  each  inch  depth 
of  metal,  or  0.00688  kilos,  per  square  centimeter  for  each  centi- 
meter depth. 

The  slag  lying  on  the  metal  has  a  specific  gravity  melted 
of  approximately  half  that  of  the  metal.  Its  amount  may  vary 
from  5  to  10  per  cent,  of  the  weight  of  the  metal  treated,  in  an 
acid-lined  converter,  up  to  from  15  to  35  per  cent,  in  a  basic- 
lined  vessel.  Taking  into  account  its  lower  specific  gravity, 
its  depth  in  the  converter  may  be,  therefore  10  to  20  per  cent, 
the  depth  of  metal  in  an  acid-lined  vessel,  and  30  to  70  per 
cent,  in  a  basic-lined  converter;  but  the  static  pressure  exerted 
would  be  only  in  direct  pioportion  to  the  relative  weights; 
i.  e.,  5  to  10  or  15  to  35  per  cent,  of  that  exerted  by  the  metal. 
The  static  pressure  of  the  slag  may,  therefore,  be  reckoned  as 
0.125  pounds  per  square  inch  for  each  inch  in  depth,  or  0.00344 


THE  BESSEMER  PROCESS.  341 

kilos,  per  square  centimeter  for  each  centimeter  depth,  and  the 
depth  of  slag  as  lying  between  the  following  extremes : 

Acid  Lined.         Basic  Lined. 

f       .,       finches..  12  to    24  12  to  24 

Depth  of  metal      |  Centimeters..      30  to    60  30  to  60 

^      ,      ,   .  /Inches 1.2  to    4.8  3.6  to  16.8 

\  Centimeters. .      3.0  to  12.0  9.0  to  42.0 

The  probable  depth  of  slag  can  be  calculated  in  any  par- 
ticular case,  when  the  composition  of  the  metal  to  be  blown  is 
known,  its  approximate  depth  in  the  vessel,  and  the  approxi- 
mate composition  of  the  slag  to  be  formed. 

The  back  pressure  of  gases  in  the  converter  itself,  that  is, 
their  static  pressure,  will  vary  with  the  shape  of  the  converter 
and  the  size  of  the  free  opening  for  their  escape  into  the  air. 
A  measurement  at  the  Pennsylvania  Steel  Go's,  works  gave 
0.275  pounds  per  square  inch,  but  it  is  not  stated  just  how 
the  measurement  was  made.  If  we  know  approximately  the 
volume  of  gas  which  must  escape  from  the  converter  and  from 
its  temperature  and  the  time  and  the  size  of  the  outlet  calculate 
its  velocity,  the  static  pressure  giving  it  this  velocity  can  be 
calculated  as 


,        V2 


in  which,  if  V  is  in  feet  per  second,  2g  =  64.3  and  the  resultant 
pressure  is  in  feet  of  the  hot  gas;  if  V  is  in  meters  per  second, 
2g  =  19.6,  and  h  is  in  meters  of  the  hot  gas.  Knowing  the 
approximate  specific  gravity  of  the  hot  gas  (weight  of  1  cubic 
foot  in  pounds  or  of  1  cubic  meter  in  kilograms)  the  static 
pressure  is  obtainable  in  pounds  per  square  foot  or  kilograms 
per  square  meter. 

Illustration:  The  gases  escaping  from  a  converter  are  78,057 
cubic  feet  (standard  conditions),  and  weigh  0.0801  pounds  per 
cubic  foot  (standard  conditions).  They  escape  from  the  con- 
verter at  an  average  temperature  of  1,500°  C.,  and  the  opening 
is  24  inches  in  diameter.  What  is  the  gaseous  back  pressure  in 
the  converter?  Time  of  blow  9  minutes  10  seconds. 


342  METALLURGICAL  CALCULATIONS. 

Volume  of  gas  at  1,500°: 

1500  +  273 


78,057  X 


273 


Volume  per  second: 

506,940^-550 

Area  of  outlet : 

2X2X0.7854 
Velocity  (assuming  0.9  coefficient) 

921.7-^0.9  +  3.1416 

Head  of  hot  gas  giving  velocity: 


h  = 


326X326 
64.3 


=  506,940  ft3 

=      921.7    " 

=     3.1416    ft2 

=         326  ft.  per  second 

=      1,653  ft. 


Pressure  of  this  column  per  square  foot : 

=        20.4  pounds 


070 
1.653X^X0.0801 


1773 


Per  square  inch 


0.14 


This  solution  omits  one  consideration;  the  velocity  of  the 
gases  in  the  body  of  the  converter  is  neglected.  This  is  some- 
what counterbalanced  by  the  great  friction  of  the  gases  against 
the  sides  of  the  converter,  so  that  the  one  item  tends  to  neu- 
tralize the  other.  If  the  interior  were  8  feet  in  diameter,  the 
velocity  of  the  gases  therein  would  average  only  some  20  feet 
per  second,  showing  the  above  corrections  to  be  practically 
negligible,  since  the  pressure  thus  represented  would  be  only 
0.4  per  cent,  of  the  total  obtained  above. 

The  pressure  necessary  to  force  the  blast  through  the  tuyeres 
is  calculable  on  principles  similar  to  the  above;  the  differences 
are  that  the  blast,  at  temperatures  varying  from  100°  C.  in  the 
blast  box  to  possibly  200°  at  its  entrance  into  the  metal,  is 
divided  up  into  fifty  or  150  streams  of  approximately  1  centi- 
meter (0.4  inch)  in  diameter,  the  length  of  tuyeres  being  some 
50  centimeters  (20  inches).  The  formula  similar  to  that  used 


THE  BESSEMER  PROCESS.  343 

for  chimney  draft,  or  rather,  friction  al  resistance  in  a  chimney, 
applies  to  this  case. 

V2  KL 


in  which  h  is  the  head  in  terms  of  the  air  passing,  V  is  its  velocity, 
2g  the  gravitation  constant,  L  the  length  of  the  tuyere,  D  its 
diameter,  and  K  the  coefficient  of  friction,  which  latter  is  for 
relatively  smooth  flues  0.05  (Grashof),  and  may  be  so  assumed 
here. 

Problem  66. 

In  the  converter  mentioned  in  Problem  65,  where  22,500 
pounds  of  metal  was  blown  in  9  minutes  10  seconds,  using 
as  therein  calculated  85,095  cubic  feet  of  air,  at  36°  C.,  and 
producing  1,374  pounds  of  slag,  assume  the  inside  diameter  of 
the  converter  as  7  feet,  and  that  the  bottom  contains  fourteen 
tuyere  blocks,  each  containing  eleven  openings  of  0.5  inch 
diameter  each;  blocks  24  inches  long.  Assume  back  pressure 
in  converter  0.14  pounds  per  square  inch,  total  blast  pressure 
in  equalizing  reservoir  27  pounds  per  square  inch.  Temper- 
ature of  air  in  the  tuyeres  150°  C. 

Required:  (1)  The  pressure  needed  to  overcome  the  head 
of  metal  and  slag. 

(2)  The  pressure  absorbed  in  friction  in  the  tuyeres. 

(3)  The  pressure  represented  by  the  velocity  of  the  blast 
in  the  tuyeres. 

(4)  The  loss  of  pressure  from  the  reservoir  to  the  blast-box. 

(5)  The  distribution  of  the  total  pressure. 

(6)  The  length  of  the  blow  if  the  blast  pressure  were  reduced 
to  20  pounds. 

(7)  The  length  of  the  blow  if  the  pressure  were  maintained 
at  27  pounds,  but  twenty-one  tuyere  blocks  (each  with  eleven 
J-inch  holes)  were  used. 

Solution:  (1)  At  the  start  there  is  22,500  pounds  of  melted 
metal,  the  volume  of  which  will  be 

22,500  22,500 

52.3  cubic  feet 


6.88X62.5  430 


344  METALLURGICAL  CALCULATIONS. 

The  depth  of  metal,  the  inside  diameter  being  7  feet,  is 

CO     O  CO     O 

bZ'6  =     1.356  feet 


7X7X0.7854       38.5 

=  16.4  inches 

Static  pressure  =  16. 4  X  0.25  =     4.1  Ibs.  per  sq.  in. 
The  slag,  formed  during  the  first  half  of  the  blow,  weighs 
1,374  pounds,  and  has  a  volume  of 

1374  1374 

=    6.4  cubic  feet 


3.44X62.5        215 

The  depth  of  slag,  at  its  maximum,  will  be 

6.4-38.5  =  0.167  feet 

=  2.0      inches 

Static  pressure  2.0X0.125       =  0.25    Ibs.  per  sq.  in. 

The  static  pressure  during  the  blow  will,  therefore,  be  4.1 

pounds  to  start  with,  increasing  during  the  first  half  of  the 

blow  to  4.35  pounds,  and  staying  practically  constant  at  that, 

and,  therefore,  will  average 

,4.1  +  4.35  ,  4.35 
2X2      +^~ 

(2)  Each  of  the  14X11  =  154  tuyeres  receives  85,095  -r- 
154-:-550  =  1.005  cubic  feet  of  air  per  second,  measured  at 
36°  C.  At  150°  C.  this  volume  is 


1.375  cubic  feet 
And  the  velocity  in  the  tuyere: 

Q-K     0.5X0.5X0.7854          ^_ 
1.375-5  --  —  -  =  1009  feet  per  second 

The  head  absorbed  in  friction  in  the  24-inch  tuyeres  will  be 

,         1009X1009^0.05X2 
"         --  X 


Changing  this  pressure  of  air  at  150°  C.  to  pounds  per  square 
inch  we  have: 


THE  BESSEMER  PROCESS.  345 

Weight  of  1  cubic  foot  of  air  at  0°              =  0.0808  pounds 

Weight  of  1  cubic  foot  of  air  at  150°          =  0.0522     " 

Weight  of  air  column  =  37,893X0.0522    =  1978     " 

Pressure  in  pounds  per  square  inch             =  13.7                     (2) 

(3)  The    pressure    absorbed    as    velocity  has    already    been 
expressed  in  getting  the  friction  in  the  tuyeres.     The  velocity 
head  is  simply: 

V2       1009X1009       1 
h  =  2F          64.3          =  1 

which  becomes  in  pressure 

15,835X0.0522  =  826.6    pounds  per  square  foot 

=       6.73  pounds  per  square  inch  (3) 

(4)  The  remaining  part  of  the  27  pounds  pressure  used  is 
lost  between  the  blast  reservoir  and  the  entrance  to  the  tuyeres. 
It  is 

27.00- (13.70  +  5.73  +  4.29  +  0.14).   =  3.14  pounds.  (4) 

(5)  Distribution  of  blast  pressure: 

Fall  between  reservoir  and  tuyeres  =   3.14  pounds  =  11.6% 

Absorbed  in  friction  in  the  tuyeres  =13.70        "        =  50.7% 

Absorbed  in  velocity  in  the  tuyeres  =   5.73       "        =  21.2% 

Static  head  of  liquid  bath  =4.29       "  15.9% 

Velocity  head  of  issuing  gases  =   0.14       "        =  0.6% 


27.00  =100.0% 

(6)  All  the  items  of  absorption  of  pressure  are  proportional 
to  the  square  of  the  velocity  of  the  gases,  excepting  the  static 
pressure  of  the  bath.  It  remains  constant  at  4.29  pounds.  If 
the  total  pressure  were  reduced  to  20  pounds,  there  would  be 
only  20—  4.29  =  15.71  pounds  pressure  to  give  velocity  and 
overcome  friction,  instead  of  27—  4.29  =  22.71  pounds.  The 
relative  quantities  of  air  blown  through  in  a  given  time  in  the 
two  instances  would  be  practically  proportional  to  the  square 
roots  of  the  two  effective  pressures,  i.e.'. 

\/22Tl  :  VIsTTl  =  1  :  0,832 
and  the  times  of  the  blows  inversely  as  the  latter: 

550  sec.  4- 0.832  =  673  seconds 

=  11  min.  13  sec.  (6) 


346  METALLURGICAL  CALCULATIONS. 

(7)  If  the  tuyere  area  were  increased  50  per  cent.,  then  the 
velocity  of  the  air  in  the  tuyeres  would  be  decreased  one-third, 
assuming  the  amount  of  air  passing  to  be  unchanged.  This 
would  decrease  the  pressure  absorbed  in  friction,  and  in  giving 
velocity  in  the  tuyeres  to  (0.67)2  =  0.444  of  its  former  amount. 
The  19.43  pounds  previously  absorbed  in  these  two  items 
would  then  become  19.43X0.444  =  8.61  pounds,  and  the  total 
pressure  needed  to  run  the  converter  just  as  fast  as  before 
would  be  27-  (19.43-  8.61)  =  16.18  pounds  per  square  inch. 
If,  however,  the  pressure  were  maintained  at  27  pounds,  giving 
still  27—  4.29  =  22.71  pounds  to  overcome  f national  resist- 
ances and  to  give  velocity,  then  the  velocity  and  consequent 
amount  of  air  blown  through  by  this  22.71  pounds  pressure 
would  increase  in  proportion  to  the  square  root  of  these  two 
available  pressures;  i.e.,  be  as 

Vl6.18-4.29  :  \/27-  4.29  =  1  :  1.38 
The  duration  of  the  blow  would  be  just  that  much  shorter ;  i.e. : 

550  sec.  -s-1.38  =  398  seconds 

=       6min.38sec.  (7) 

FLUX  AND  SLAG. 

No  flux  is  used  in  the  acid-lined  converter,  and  the  silica, 
iron  oxides  and  manganese  oxide  formed  in  the  converter 
unite  to  a  silicate  slag  which  corrodes  the  lining  and  thus  takes 
up  more  silica.  The  slag  being  analyzed,  its  weight  is  ob- 
tained by  considering  the  percentage  of  manganese  which  it 
contains,  because  the  weight  of  manganese  oxidized  is  known 
definitely  from  the  analysis  of  the  bath;  it  is  usually  all  oxi- 
dized. Calculation  of  the  weight  of  slag  cannot  be  based  upon 
the  silica,  because  an  unknown  amount  comes  from  the  lining ; 
nor  upon  the  iron,  because  the  weight  of  iron  left  in  the  con- 
verter is  not  definitely  known.  Having  the  weight  of  the  slag, 
analysis  tells  us  the  total  weight  of  silica  in  it,  as  also  the  amount 
of  iron.  The  silica  in  the  slag  minus  that  formed  from  silicon 
in  the  pig  iron,  gives  silica  corroded  from  the  lining. 

In  the  basic  Bessemer  converter,  phosphorus  is  nearly  en- 
tirely eliminated  from  the  metal,  so  that,  assuming  none  to  be 
volatilized,  the  amount  going  into  the  slag  is  known,  and  using 
the  slag  analysis  the  weight  of  slag  can  be  calculated.  In 


THE  BESSEMER  PROCESS.  347 

this  process  the  lining  is  mainly  dolomite,  containing  CaO  and 
MgO,  in  proportion  easily  determined  by  analysis.  The  weight 
of  slag  being  known,  the  amount  of  corrosion  of  the  lining 
can  be  determined  from  the  percentage  of  magnesia  therein, 
which  may  be  assumed  as  practically  coming  entirely  from 
the  lining ;  it  cannot  be  told  from  the  CaO  in  the  slag,  because 
nearly  pure  CaO  is  added  during  the  blow,  and  some  of  it, 
a  variable  amount,  gets  blown  out  of  the  converter.  For 
the  same  reason  it  is  not  possible  to  base  a  good  calculation 
of  the  weight  of  slag  on  the  lime  alone  which  is  added,  be- 
cause of  the  indefinite  proportion  of  it  which  is  blown  out. 
The  weight  of  slag  may  also  be  gotten  from  the  silica  or  man- 
ganese oxide  in  it,  assuming  these  to  come  almost  entirely  from 
the  oxidation  of  silicon  or  manganese. 

Lime  must  be  added  as  flux,  in  the  basic  converter,  to  pro- 
tect the  lining  and  to  make  the  slag  so  basic  that  the  percentage 
of  silica  in  it  is  below  15  per  cent.,  phosphoric  acid  below  20 
per  cent.,  and  lime  over  50  per  cent.  These  considerations  must 
be  balanced  in  each  particular  case. 

Illustration:  Pig  iron  blown  in  a  basic-lined  converter  con- 
tained 1.22  per  cent,  silicon,  2.18  phosphorus,  1.03  manganese 
and  3.21  carbon.  It  is  blown  until  all  of  these  and  2.00  per 
cent,  of  iron  are  oxidized,  and  burnt  lime  is  added  to  form 
slag  during  the  blow.  Composition  of  the  burnt  lime:  MgO, 
1.00  per  cent.;  SiO2,  2.00  per  cent.;  CaO,  97  per  cent.  How 
much  lime  should  be  added  per  10  metric  tons  of  pig  iron  charged? 

The  slag-forming  ingredients  from  the  oxidation  of  the  bath, 
and  the  addition  of  X  kilos,  of  lime,  are 

SiO2  10,000X0. 0122 X  60/28  =  261.4  kg. 

P2O5  .    10,000X0.0218X142/62  =  499.3    " 

MnO  10,000X0. 0103 X  71/55=133.0    " 

FeO  10,000X0. 0200 X  72/56=257.1    " 

fCaO  XX 0.9700  =    0.97  X 

\  MgO  XX 0.0100  =    0.01  X 

[SiO2  XX 0.0200  =    0.02  X 


Weight  for  slag  =  X  + 1150.8  kg. 

Corrosion  of  the  lining  will  undoubtedly  increase  this  weight, 
so  some  allowance  should  be  made,  say  to  increase  it  5  per 


348  METALLURGICAL  CALCULATIONS. 

cent.,  probably  an  outside  figure.  Of  this  5  per  cent.,  half  can 
be  considered  lime  and  half  magnesia.  The  total  weight  of 
slag  will  then  be  1.05  X-f  1208.3,  and  of  the  ingredients  prin- 
cipally in  question: 

SiO2  =  261.  kg. +  0.02  X 
MgO  =  28.7  "  +0.035  X 
CaO  =  28.7  "  +  0.995  X 

To  make  our  slag  50  per  cent.  CaO  will  require  the  addition 
of  enough  to  make 

28. 7  +  0. 995  X  =  0.50  (1.05  X  +  1208.3) 
X  =  1224  kg. 

To  make  a  slag  with  at  most  15  per  cent,  of  SiO2  requires 

261.4  +  0.02  X  =  0.15  (1.05  X  +  1208.3) 
X  =    583  kg. 

To  make  a  slag  with  at  most  20  per  cent,  of  P2O5  requires 

499.3  =  0.20  (1.05  X  +  1208.3) 
X  =  1227  kg. 

The  larger  of  these  three  amounts  would  be  used,  with  10 
per  cent,  added  to  cover  lime  dust  blown  out,  making  1350  kg. 
added,  and  the  calculated  composition  of  the  slag: 

CaO  1250  kg.  =  50.1  per  cent. 

MgO  723    "    =     2.9 

SiO2  286   "    =  11.4 

P2O5  499   "    =  20.0 

FeO  257    "    =  10.3 

MnO  133    "    =    5.3 


Total  2497 

RECARBURIZATION. 

When  the  bath  has  been  blown  to  nearly  pure  iron,  melted 
spiegeleisen  is  run  in,  to  add  the  necessary  carbon  and  man- 
ganese. Knowing  the  approximate  composition  and  weight  of 
the  bath,  and  the  composition  of  the  melted  Spiegel,  a  simple 
arithmetical  calculation  would  give  the  amount  of  the  latter 
to  be  added,  assuming  no  loss  of  carbon  or  manganese  in  the 


THE  BESSEMER  PROCESS.  349 

operation.  But  experience  shows  that  there  is  some  loss,  and 
that  the  carbon  and  manganese  in  the  finished  metal  are  always 
lower  than  the  calculated  amount.  An  interesting  field  is  open 
here  for  calculating  the  loss  of  manganese  and  carbon  and  the 
amount  of  oxygen  which  must  have  been  in  the  metal  to  cause 
these  losses.  A  tabulation  of  many  such  calculations  gives 
the  metallurgist  the  necessary  data  for  assuming  the  average 
amounts  of  carbon  and  manganese  lost  during  recarburization, 
under  different  conditions  of  working,  such  as  letting  the  metal 
stand  before  pouring  or  pouring  at  once,  turning  on  the  blast 
5  or  10  seconds  to  mix  up  the  bath,  etc. 

Problem  67. 

At  the  end  of  the  blowing  the  converter  of  Problem  65  con- 
tained 21,283  pounds  of  metal  of  the  composition  0.04  per 
cent,  carbon,  0.02  silicon,  0.01  manganese,  0.11  phosphorus,  0.06 
sulphur,  an  unknown  amount  of  oxygen  (probably  <  0.3  per 
cent.)  and  the  rest  iron.  There  is  added  to  it  2,500  pounds 
of  spiegeleisen,  containing  4.64  per  cent,  carbon,  0.035  silicon, 
14.90  manganese,  and  0.139  phosphorus.  The  finished  metal 
contained  0.45  per  cent,  of  carbon,  0.038  silicon,  1.15  manganese, 
0.109  phosphorus  and  0.059  sulphur.  Assume  no  iron  oxidized. 
Required:  (1)  A  balance  sheet  of  materials  before  and  after 
recarburizing. 

(2)  The  proportions  of  carbon  and  manganese  going  into  the 
finished  metal. 
Solution:     (1) 

Blown  Gases 

Metal     Spiegel.        Steel.      or  Slag. 

C 8.5         116.0         106.5  18.0 

Si 4.3  0.9  9.0          -3.8 

Mn 2.1         372.5         271.2  103.4 

P 23.4  3.5  25.8  1.1 

S 12.8  ....  14.0          -  1.2 

Fe 21,232       2007          23,239 

The  differences  in  the  sulphur,  phosphorus  and  silicon  are 
within  the  limits  of  error  of  the  data,  but  there  is  no  doubt  as 
to  the  loss  of  carbon  and  manganese. 

(2)  The  proportions  of  the  two  elements  in  question  going 
into  the  finished  steel  are: 


350  MET  A  LL  URGICAL  CALCULA  TIONS. 

Carbon 106.5^-124.5  =  0.85  =  85  per  cent. 

Manganese 271.2-^-374.6  =  0.72  =  72 

The  calculated  percentages  in  the  finished  steel  should  have 
been,  and  actually  were: 

Carbon 0.53-  0.45,  loss  =  0.08  per  cent. 

Manganese 1.58-  1.15,  loss  =  0.43 

Concerning  oxygen  removed,  if  we  assume  the  loss  of  carbon 
and  manganese  to  be  due  to  their  combining  with  oxygen  dis- 
solved in  the  bath,  to  form  CO  and  MnO,  the  percentage  of 
oxygen  thus  absorbed  is: 

By  carbon 0.08x16/12  =  0.11  per  cent. 

By  manganese 0.43X16/55  =  0.12 

0.23 

[The  next  chapter  will  consider  the  thermo-chemistry  of  the 
Bessemer  process.] 


CHAPTER  VIH. 
THERMOCHEMISTRY   OF   THE    BESSEMER    PROCESS. 

The  feature  of  the  Bessemer  operation  which  strikes  the 
observer  as  most  wonderful,  is  that  cold  air  is  blown  in  great 
quantity  through  melted  pig  iron,  and  yet  the  iron  is  hotter  at 
the  end  than  at  the  beginning.  If  the  observer  will  reflect  a 
moment,  however,  he  can  see  that  if  nothing  but  fuel,  on  fire, 
was  in  the  converter,  it  would  certainly  be  made  much  hotter 
by  the  air  blast ;  in  similar  manner,  the  oxidation  or  combustion 
of  part  of  the  ingredients  of  the  pig  iron  furnishes  all  the  heat 
required  for  the  process.  Ten  tons  of  pig  iron  contains,  for 
example,  some  350  kilograms,  of  carbon  which  is  all  burnt  out 
in  the  bussemerizing,  furnishing  heat  equal  to  the  combustion 
of  some  400  kilograms  of  coke — a  not  insignificant  quantity, 
since  it  is  burned  and  its  heating  power  utilized  in  a  very  few 
minutes. 

ELEMENTS  CONSUMED. 

The  usual  ingredients  of  pig  iron  are: 

Iron 90.0  to  95.0    per  cent 

Carbon 2.5  "  4.5 

Manganese 0.5  "  4.0 

Silicon 0.5  "  4.0 

Phosphorus 0.01  "  4.0 

Sulphur 0.01  "  0.5 

Some  of  the  unusual  constituents  are  nickel,  chromium,  titan- 
ium, aluminium,  vanadium,  tungsten,  copper  and  zinc;  all  of 
these  are  rare,  and  there  is  seldom  present  as  much  as  0.5  per 
cent,  of  any  one  except  in  unusual  cases. 

In  the  Bessemer  operation,  carried  out  with  the  usual  silica 
lining,  iron,  carbon,  manganese  and  silicon  are  freely  oxidized, 
but  phosphorus  and  sulphur  remain  practically  unchanged.  In 
the  basic  Bessemer,  lined  with  burnt  dolomite  and  tar,  phos- 

351 


352  METALLURGICAL  CALCULATIONS. 

phorus  is  also  freely  oxidized  at  the  end  of  the  operation,  but 
sulphur  is  only  slightly  diminished — the  more,  the  more  man- 
ganese is  in  the  slag.  After  all  the  oxidizable  impurities  are 
eliminated,  iron  itself  oxidizes  in  much  larger  quantity,  oc- 
casioning great  loss  if  the  blast  is  permitted  to  continue. 

Iron  oxidizes  during  the  blow  mostly  to  FeO,  which  enters 
the  slag  as  ferrous  silicate,  and  partly  to  Fe203.  The  brown 
fume  which  escapes  in  large  amount  if  the  blow  is  continued 
too  long  contains  iron  as  Fe203. 

Fe  to  FeO 1173  Calories  per  kg.  bf  Fe 

Feto  Fe2O3 1746 

FetoFe304 1612 

The  amount  of  iron  oxidized  can  be  gotten  from  the  weight 
and  percentage  composition  of  the  slag;  also  from  the  com- 
parison of  the  weights  of  materials  used  and  weight  of  ingots 
produced,  knowing  the  weight  of  other  impurities  oxidized. 

Carbon  oxidizes  mostly  to  CO  gas,  and  partly,  especially  in 
the  early  part  of  the  blow,  to  CO2  gas.  The  proportion  of  each 
of  these  formed  can  only  be  known  by  analyzing  the  gases 
produced  at  various  stages  of  the  blow.  The  proportionate 
volumes  of  CO  and  CO2  express  the  proportionate  amounts 
of  carbon  forming  each  respective  gas.  A  shallow  bath  allows 
more  CO2  to  pass,  on  essentially  the  same  principle  that  a  deep 
layer  of  fuel  on  a  grate  favors  the  production  of  CO.  The 
heat  evolved  by  oxidation  of  carbon  is : 

C  to  CO 2430   Calories  per  kg.  of  C 

CtoCO2 8100 

Manganese  oxidizes  quickly  and  mostly  to  MnO.  If  the 
metal  is  a  little  overblown,  Mn2O3  in  small  amount  is  found  in 
the  slag,  while  Mn2O3  is  also  present  in  the  fumes.  The  heat 
evolved  in  these  oxidations  is : 

Mn  to  MnO 1653  Calories  per  kg.  of  Mn 

MntoMn2O3 2480  (?)  " 

The  last  figure  is  estimated;  it  has  not  yet  been  determined 
experimentally. 

Silicon  oxidizes  rapidly  and  early  in  the  blow  to  SiO2,  form- 
ing silicate  slag  with  the  metallic  oxides  formed.  Its  heat 


THE  BESSEMER  PROCESS.  353 

of  oxidation  has  been  usually  taken  as  7830  Calories  per  kilo- 
gram, but  recent  investigations  have  thrown  doubt  on  this 
figure,  Berthelot  having  found  as  low  as  6414  Calories,  and  Mr. 
H.  N.  Potter,  7595  for  crystalline  silicon,  equal  to  7770  for 
amorphous  silicon.  Under  these  circumstances  the  best  course 
is  probably  to  use  the  middle  value  ad  interim,  and  consider 

Si  to  SiO2 7000  Calories  per  kg.  of  Si 

We  hope  that  the  exact  figure  will  soon  be  determined. 

Phosphorus  oxidizes  to  P2O5,  and  only  towards  the  end  of 
the  blow.  It  is  practically  completely  eliminated,  going  into 
the  slag  as  calcium  phosphate : 

P  to  P205 5892  Calories  per  kg.  of  P 

Sulphur  is  not  eliminated  at  all  in  the  acid-lined  converter. 
In  the  basic  converter  it  is  reduced  in  amount  in  the  last  few 
minutes,  while  phosphorus  is  disappearing,  and  partly  escapes, 
mostly  as  SO2  in  the  gases.  The  presence  of  a  very  basic  slag 
is  necessary,  but  the  sulphur,  while  possibly  going  into  the  slag, 
does  not  remain  there,  but  passes  into  the  gases.  The  heat  of 
oxidation  of  the  unusual  elements  sometimes  present  are,  as 
far  as  known, 

Ni  to  NiO 1051  Calories  per  kg.  of  Ni 

Ti  to  TiO2 4542       "  "  Ti 

Al  to  A12O3 7272      "  "  Al 

ZntoZnO 1305       "  "  Zn 

Vto  V2O5 4324  .    "  V 

^WtoWO3.... 1047       "  "  W 

CrtoCrW 2344      "  "  Cr 

Some  of  the  above  values  are  a  little  uncertain,  and  none  of 
them  include  the  heat  of  combination  of  the  oxide  formed  with 
the  slag. 

HEAT  BALANCE  SHEET. 

Taking  o°  C.  (32°  F.)  as  a  base  line,  we  may  express  the 
total  heat  contents  of  the  pig  iron,  steel,  gases,  slag,  blast,  etc., 


354  METALLURGICAL  CALCULATIONS. 

from  this  temperature.     This  method  is  simpler  than  to  take  the 
bath  at  any  one  high  temperature  and  to  reckon  from  there. 
The  items  of  heat  available  during  the  blow  are: 

Heat  in  the  body  of  converter  at  starting. 

Heat  in  the  melted  pig.  iron  used. 

Heat  in  the  spiegleisen  or  ferro-manganese  added. 

Heat  in  hot  lime  added  (sometimes  in  basic  process). 

Heat  in  the  blast,  if  warm  on  entering. 

Heat  developed  by  oxidation  of  the  bath. 

Heat  developed  by  formation  of  the  slag. 
The  items  of  heat  distribution  are: 

Heat  in  the  body  of  the  converter  at  finishing. 

Heat  in  the  steel  poured. 

Heat  in  the  slag  at  finishing. 

Heat  in  the  gases  escaping. 

Heat  in  the  fume. 

Heat  in  the  slag  or  metal  blown  out. 

Heat  absorbed  in  decomposing  moisture  of  the  blast. 

Heat  to  separate  the  constituents  of  the  bath. 

Heat  conducted  away  by  supports,  blast  pipe,  etc. 

Heat  conducted  to  the  air  in  contact  with  converter. 

Heat  radiated  during  the  blow. 

These  two  columns  should  balance  each  other  if  all  the  items 
of  each  are  correctly  determined. 

Heat  in  Converter  Body  at  Starting. 

If  the  converter  were  quite  cold  when  the  pig  iron  was  run 
in  and  the  blow  started,  this  item  would  be  zero.  But  such  a 
case  would  result  disastrously,  since  the  heat  absorbed  by  the 
converter  during  the  blow  would  be  more  than  any  ordinary 
heat  could  afford  to  lose.  It  is,  therefore,  customary,  when 
starting  for  the  first  time,  to  build  a  fire  in  the  converter  and 
turn  on  a  little  air  blast,  so  as  to  bring  the  inside  up  to  bright 
redness,  say  900°  to  1000°  C.  The  outside  shell  would,  under 
these  conditions,  be  at  about  200°,  and  the  mean  temperature  of 
the  converter  lining,  say  400°.  If  the  converter  were  in  regular 
operation,  one  charge  being  introduced  as  soon  as  the  other 
was  finished,  then  the  heat  in  the  body  of  the  converter  at  start- 
ing could  be  practically  regarded  as  equal  to  the  heat  in  the 


THE  BESSEMER  PROCESS.  355 

same  at  finishing,  assuming  the  heats  are  running  regularly 
An  estimate  of  the  heat  in  the  converter  body  at  starting  is 
therefore  only  necessary  when  the  converter  is  first  started  up, 
or  when  it  is  allowed  to  stand  some  time  between  blows. 

Illustration:  Assume  a  converter  weighing,  without  charge, 
25  tons,  of  which  5  tons  is  iron  work  and  20  tons  silica  lining. 
The  mean  specific  heat  of  iron  (for  low  temperatures)  being 
0.1 1012 +  0.000025t  +  0.0000000547t2,  and  for  silica  0.1833  + 
0.000077t,  calculate  the  heat  contained  in  the  body  of  the  con- 
verter.— 

(1)  When  the  temperature  of  the  outside  shell  is    200°  and 
that   of  the  lining  averages   400°    (converter  warmed  up  for 
starting). 

(2)  When  the  temperature  of  the  outside  shell  is  300°  and 
that  of  the  lining  averages  725°  (converter  empty  at  end  of  a 
blow) . 

Solution:     (1) 

Heat  in  5000  kg.  of  iron  work: 

0.11012  +  0.000025  (200) +0.0000000547  (200)2  =  0.11731 
0.11731X200X5000  =  117,310  Calories. 

Heat  in  20,000  kg.  of  silica  lining: 
0.1833  +  0.000077  (400)  =       0.2141 

0.2141X400X20,000  =  1,712,800  Calories. 

Total  heat  in  converter  body  =  1,830,110 

It  may  be  remarked  that  this  would  require  the  consumption 
of  at  least  250  kilograms  of  coke,  with  a  calorific  power  of 
8,000  Calories,  to  warm  the  converter  to  this  extent. 

(2)  Heat  in  5,000  kg.  of  iron  work: 

0.12354X300X5,000          =     185,310  Calories. 
Heat  in  20,000  kg.  of  silica  lining: 

0.2391   X  725X20,000         =3,466,950 


Total        =  3,652,260 

This  condition  is  assumed  as  representing  the  .converter  when 
just  emptied  and  immediately  refilled.  In  this  case,  in  regular 
working,  the  heat  in  the  converter  body  is  practically  the  same 
at  the  beginning  and  at  the  end  of  a  blow,  and  the  heat  losses 
through  it  are  only  those  due  to  conduction  to  the  air  and 
ground  and  radiation. 


356  METALLURGICAL  CALCULATIONS. 

Heat  in  Melted  Pig  Iron  Used. 

This  quantity  depends  on  the  temperature  at  which  the  pig 
iron  is  run  into  the  converter.  If  the  iron  is  high  in  silicon, 
which  would  tend  to  produce  a  hot  blow,  it  may  be  run  in 
somewhat  cool;  but  if  low  in  silicon  it  should  be  run  in  much 
hotter.  The  minimum  quantity  of  heat  contained  in  a  kilogram 
of  melted  pig  iron  may  be  put  at  250  Calories;  the  maximum, 
for  very  hot  pig  iron,  350  Calories;  about  300  Calories  would 
be  a  usual  average  figure.  This  may  be  easily  determined  ex- 
perimentally in  any  given  case  by  granulating  a  sample  in 
water  in  a  rough  calorimeter. 

Heat  in  Metallic  Additions. 

Melted  spiegeleisen  is  usually  not  very  hot  when  run  into  the 
converter.  It  may  contain  250  to  300  Calories  per  kilogram; 
say  an  average  of  275.  Ferro-manganese  is  sometimes  added 
red-hot,  at  800°  to  900°  C.  At  this  temperature  it  would  con- 
tain 120  to  135  Calories  per  kilogram,  assuming  a  mean  specific 
heat  of  0.15. 

Heat  in  Preheated  Lime. 

Taking  the  mean  specific  heat  of  CaO  as  0.1715  +  0.00007t, 
it  would  contain  154  Calories  per  kilogram  at  700°,  and  211 
Calories  at  900°.  The  heat  content  can  be  calculated  for  any 
known  temperature  at  which  the  lime  is  used. 

Heat  in  Warm  Blast. 

No  Bessemer  converters  are  run  by  hot  blast,  but  the  air 
pressure  used  is  so  great  (20  to  35  pounds  per  square  inch) 
that  the  blast  is  warmed  by  compression  in  the  cylinders.  If 
air  at  1  atmosphere  tension  (ordinary  air)  be  compressed  to  2 
or  to  3  atmospheres  tension,  giving  effective  blast  pressures  of 
1  to  2  atmospheres,  the  air  is  heated  60°  or  103°,  respectively, 
above  its  initial  temperature.  While  some  of  this  heat  may  be 
lost  in  the  cylinder  and  conduits,  yet  the  air,  unless  artificially 
cooled,  passes  to  the  tuyeres  at  25°  to  50°  C.  above  the  outside 
temperature,  and  thus  imparts  some  heat  to  the  converter. 

Heat  Developed  by  Oxidation. 

We  have  already  discussed  the  thermochemical  data  required 
for  this  calculation.  To  use  the  data  we  must  find  the  weights 


THE  BESSEMER  PROCESS.  357 

of  each  ingredient  oxidized  (not  forgetting  the  iron  itself)  and 
the  nature  of  the  oxide  it  forms.  This  is  deduced  from  the 
analysis  of  slag,  gases  and  steel  produced,  as  compared  with 
those  of  the  pig  iron  and  additions  used. 

Heat  of  Formation  of  Slag. 

The  slag  is  a  mechanical  mixture  or  mutual  solution  of  sili- 
cates of  iron  and  manganese,  with  a  little  alumina  (up  to  5 
per  cent.)  and  lime  and  magnesia  from  nothing  up  to  5  per 
cent.  In  the  basic  process  a  large  amount  of  calcium  phos- 
phate is  present,  representing  over  half  of  the  entire  slag,  while 
magnesia  is  present  in  considerable  amount,  and  alumina  is 
almost  absent. 

Having,  in  the  preceding  column,  calculated  the  heat  of 
oxidation  of  all  the  metals  oxidized  in  the  converter,  it  re- 
mains to  calculate  the  heat  of  combination  of  these  to  form 
slag.  The  silicate  of  alumina  contributes  nothing,  since  it 
occurred  combined  in  the  lining  or  lime  added.  The  same  can 
be  said  of  the  ,lime  and  magnesia  in  the  acid  process  slags. 
The  FeO  and  MnO  are  then  to  be  considered,  and  the  only 
thermochemical  data  we  have  are: 

(MnO,  SiO2)  =  5,400  Calories. 
(FeO,  SiO2)  =  8,900 

These  data  are  for  71  or  72  parts  of  MnO  or  FeO  respectively, 
uniting  with  60  parts  of  SiO2.  Since  in  acid  slags  there  is 
always  proportionately  less  of  the  bases,  we  should  utilize 
the  above  heats  of  formation  by  expressing  them  per  unit 
weight  of  MnO  or  FeO  going  into  combination.  These  figures 
are: 

Per  kg.  of  MnO  =  5,400-^-71  =    76  Calories. 

Per  kg.  of  FeO   =  8,900^72  =  124 

From  these  figures  the  heat  of  formation  of  the  slag  can  be 
calculated:  Any  Fe2O3  present  in  the  slag  would  be  calculated 
as  its  equivalent  weight  of  FeO  (by  multiplying  by  1444-160). 
Illustration:     A  Bessemer  slag  contained  by  analysis: 

SiO2 47.25  per  cent. 

A12O3 3.45 

FeO 15.43 

MnO 31.89 

CaO,  MgO 1.84 


358  METALLURGICAL  CALCULATIONS- 

What  is  its  heat  of  formation  per  kilogram? 

Solution:  The  APO3,  CaO  and  MgO  are  to  be  neglected, 
for  reasons  already  given.  The  heat  of  combination,  per  kilo- 
gram of  slag,  is  therefore, 

FeO  uniting  with  SiO2    =  0.1543X124  =  19.1  Cal. 

MnO  uniting  with  SiO2  =  0.3189 X   76  =  23.2    " 

Total  =  42.3     " 

In  basic  Bessemer  slags  the  conditions  are  much  more  com- 
plicated. The  content  of  P2O5  is  not  under  14  per  cent.,  runs 
as  high  as  25  per  cent.,  and  averages  19  per  cent.;  the  CaO 
averages  45  per  cent.,  limits  35  to  55;  the  silica  is  usually  below 
12  per  cent.,  and  averages  6  to  8  per  cent.;  magnesia  is  present 
from  1  to  7  per  cent.,  average  about  4  per  cent.  In  such  slags 
we  should  first  of  all  assume  the  P2O5  to  be  combined  with 
CaO  as  3CaO.  P2O5,  containing  168  of  CaO  to  142  of  P2O5, 
(1,183  to  1)  and  having  a  heat  of  formation  from  CaO  and 
P205  of  159,400  Calories  per  molecule,  or  1123  Calories  per 
unit  weight  of  P2O5.  Next  the  iron  and  manganese  present 
may  be  calculated  to  FeO  and  MnO  respectively,  and  treated 
as  to  their  combination  with  SiO2,  the  same  as  in  an  acid  slag. 
Alumina  may  be  considered  in  such  basic  slags  as  an  acid,  and 
equivalent  to  120/102  of  its  weight  of  silica.  These  allowances 
will  leave  considerable  lime  and  either  an  excess  or  deficiency 
of  silica;  if  an  excess,  we  can  assume  it  combined  with  lime 
and  magnesia,  with  a  heat  evolution  equal  to  476  Calories  per 
kilogram  of  silica;  if  a  deficiency,  we  let  the  calculations  stand 
without  further  modification. 

Illustration:    Slag  made  at  a  Rhenish  works  contained: 

SiO2 7.73  per  cent. 

P2O5 21.90 

APO3 3.72 

Fe2O3 1.00 

FeO 4.73 

MnO 2.05 

CaO 50.76 

MgO 4.00 

CaS 1.71 

What  is  its  heat  of  formation  per  unit  of  slag? 


THE  BESSEMER  PROCESS.  359 

Solution:  The  21.90  parts  of  P205  would  be  combined  with 
21.90X168/142  =  25.91  parts  of  CaO.  This  leaves  50.76— 
25.91  =  24.85  of  CaO  as  either  free,  dissolved  CaO  or  partly 
combined,  in  the  slag.  The  2.05  MnO  would  correspond  to 
2.05X60/71  =  1.73  SiO2;  the  4.73  FeO  to  4.73X60/72  = 
3.94  SiO2;  the  1.00  Fe2O3  to  1.00X60/80  =  0.75  SiO2;  a  total 
SiO2  requirement  for  these  three  bases  of  6.42  per  cent.  The 
SiO2  present  is  7.73  per  cent.,  adding  to  which  the  SiO2  equiva- 
lent to  the  A12O3  present  (3.72X120/102  =  4.38),  we  have 
12.11  per  cent,  of  summated  silica.  The  ratio  of  summated 
FeO  to  summated  SiO2  is  considerably  below  the  ratio  72  to 
60,  we  can,  therefore,  consider  the  summated  FeO  as  all  com- 
bined with  silica.  The  summated  FeO  is : 

FeO  =  4.73  per  cent. 

FeO  equivalent  of  MnO  =  2.08 

FeO  equivalent  of  Fe203  =  0.90       " 

Total  =  7.71       " 
And  the  SiO2  combining  with  this  as  FeO. SiO2  is 

7.71X60/72  =  6.42  per  cent. 

The  excess  of  summated  silha  free  to  combine  with  lime  is 
12.11 — 6.42  =  5.69  per  cent.  As  there  is  24.85  of  CaO  and 
4.00  of  MgO  for  it  to  combine  with,  the  heat  of  this  combina- 
tion must  be  calculated  on  the  SiO2  going  into  this  combination. 

We  then  have  the  formation  heat  of  the  slag  as: 

P2O5  to  3CaO  .  P205 21.90  X 1123  =  24,594  Cal. 

MnO  to  MnO  .  SiO2 2.05X     76=        156    " 

FeO  to  FeO    .SiO2 5.63  X   124=       698   " 

SiO2  to  3CaO  .  SiO2 5.69  X   476  =    2,708   " 


28,156   ' 

This  equals  281.6  Calories  per  unit  weight  of  slag,  forming 
a  very  important  item  in  the  heat  balance  sheet,  particularly 
when  the  slag  is  large  in  amount. 

Heat  in  Converter  Body  at  Finishing. 

This  item  reaches  its  maximum  at  the  end  of  the  blow,  and 
would  be  equal  to  that  calculated  for  the  beginning  of  the  blow, 


360  METALLURGICAL  CALCULATIONS. 

with  the  exception  that  some  of  the  lining,  silica  or  dolomite 
has  been  corroded  and  passed  into  the  slag,  carrying  with  it  its 
sensible  heat. 

Heat  in  Finished  Steel. 

This  should  be  determined  experimentally  in  each  particular 
case.  If  not  so  determined  an  average  value  may  be  assumed, 
based  on  the  following  considerations:  The  finishing  tempera- 
ture averages  1650°  C. ;  at  this  heat  average  Bessemer  steel 
will  contain  a  total  of  350  Calories  of  heat  per  kilogram.  If  the 
temperature  is  determined  by  a  pyrometer,  a  correction  of  1/5 
Calorie  can  be  made  for  every  degree  hotter  or  colder  than 
1650°. 

Heat  in  Slag. 

Some  experimental  data  are  badly  needed,  concerning  the 
heat  in  slags  of  different  composition  at  different  temperatures. 
At  present  it  is  necessary  to  make  guesses,  wherever  the  heat 
in  the  slag  produced  is  not  directly  determined.  At  a  finishing 
temperature  of  1650°  it  is  likely  that  the  slag  contains  550 
Calories  per  kilogram;  with  a  variation  of  1/4  Calorie  for  each 
degree  hotter  or  colder  than  1650°. 

Heat  in  Escaping  Gases. 

The  amount  of  these  gases  can  only  be  determined  satis- 
factorily from  their  analysis  and  the  known  weights  of  carbon 
oxidized.  Direct  estimation  from  the  piston  displacement  is  of 
much  less  exactness,  because  the  slip  and  leakage  at  these 
high  pressures  may  reach  25  to  50  per  cent,  or  even  more.  The 
temperature  of  the  gases  is  only  slightly  less  than  that  of  the 
bath,  that  is,  some  1350°  at  starting  and  1650°  at  finishing. 
Outside  in  the  air  the  Bessemer  flame  may  be  much  hotter  than 
this,  but  that  is  due  to  further  combustion  of  CO  to  CO2  out- 
side the  converter,  and  should  be  disregarded.  Where  extra  air 
is  blown  upon  the  surface  of  the  bath,  as  in  "  baby  "  converters, 
it  is  quite  possible  that  the  gases  in  the  converter  and  the  es- 
caping gases  may  be  considerably  hotter  than  the  bath  itself 
These  variations  must  be  taken  into  consideration.  The  most 
satisfactory  condition  is  to  insert  a  pyrometer  tube  into  the 
opening  of  the  converter  and  measure  the  temperature  directly. 
The  heat  carried  out  by  the  gases  can  then  be  calculated  ac- 


THE  BESSEMER  PROCESS.  361 

curately,  using  the  proper  mean  specific  heats  to  these  high  tem- 
peratures already  used  in  these  calculations. 

Heat  in  Escaping  Fume. 

This  is  mostly  oxides  of  iron  and  manganese,  with  some- 
times silica.  It  is  relatively  small  in  amount.  Its  quantity 
being  known,  consider  it  at  the  same  temperature  as  the  gases, 
with  a  mean  specific  heat  of  0.40  if  free  from  silica,  and  0.35 
if  siliceous. 

Heat  in  Slag  or  Metal  Blown  Out. 

These  can  be  counted  as  equal  to  the  heat  in  an  equal  quantity 
of  slag  or  metal  at  the  finishing  temperature. 

Heat  Absorbed  in  Decomposing  Moisture. 
Knowing  the  amount  of  moisture  blown  in  with  the  blast,  a 
proper  allowance  is  29,040 -f- 9  =  3,227  Calories  for  every  kilo- 
gram of  moisture  thus  blown  in.  It  is  probable  that  this 
moisture  is  all  decomposed,  its  hydrogen  appearing  in  the 
gases.  In  the  absence  of  data  as  to  the  hygrometric  condition 
of  the  blast,  the  amount  of  moisture  entering  may  be  inferred 
and  calculated  from  the  amount  of  hydrogen  in  the  gases. 

Heat  to  Separate  Constituents  of  Bath. 

We  here  meet  the  question,  how  much  heat  of  combination 
exists  between  the  bath  and  the  various  ingredients  which  are 
removed — carbon,  silicon,  manganese,  phosphorus,  sulphur. 
Le  Chatelier  believes  manganese  to  exist  in  the  bath  as  Mn3C, 
requiring  80  Calories  per  kilogram  of  manganese  to  decompose 
it.  Silicon  and  carbon  have,  as  far  as  has  at  present  been 
determined,  no  sensible  heat  of  combination  with  iron.  Sulphur 
requires  750  Calories  to  separate  each  kilogram  from  iron. 
Phosphorus  requires,  according  to  Ponthiere,  1397  Calories  to 
separate  each  kilogram  from  iron,  but  the  reliability  of  this 
datum  is  doubtful.  Until  more  reliable  tests  are  made  it  is 
perhaps  better  to  omit  this  item  than  to  use  it,  although  it  must 
be  of  great  importance  if  as  large  as  Ponthiere  states  it  to  be. 

Heat  Conducted  Away  by  Supports. 

This  is  a  very  difficult  quantity  to  determine,  being  con- 
ditioned by  the  size  of  the  supports,  their  cooling  surface  and 


362  METALLURGICAL  CALCULATIONS. 

the  kind  of  connection  they  have  with  other  objects.  The  heat 
which  would  pass  to  the  blast  pipe  is  practically  returned  to  the 
converter  by  the  incoming  blast.  The  heat  passing  into  the 
supports  is  perhaps  best  found  by  taking  their  temperature  at 
different  places,  and  calculating  the  heat  loss  from  their  sur- 
face by  radiation  and  conduction  to  the  air.  This  amounts 
practically  to  considering  them  as  part  of  the  outer  cooling 
surface  of  the  converter;  the  calculation  of  these  surface  losses 
is  given  in  the  two  following  paragraphs: 

Heat  Conducted  to  the  Air. 

This  is  a  function  of  the  extent  of  outside  surface,  its  tem- 
perature, the  temperature  of  the  air,  and  the  velocity  of  the 
air  current.  Measurement  will  give  the  extent  of  surface  in 
contact  with  the  air  and  the  average  velocity  of  the  air  cur- 
rent; the  surface  being  rough  iron,  the  coefficient  of  transfer 
conductivity  may  be  taken  as  k  =  0.000028  (2  +  V~v).  where 
v  is  the  air  velocity  in  c.m.  per  second,  and  k  is  the  heat  con- 
ducted per  second  in  gram-calories,  from  each  square  c.m. 
of  surface  per  1°  difference  of  temperature.  The  temperature 
of  the  outer  surface  should  be  carefully  measured,  so  that  a 
reliable  average  is  obtained,  and  the  air  velocity  likewise  aver- 
aged, since  it  has  considerable  influence  on  the  heat  lost  to 
the  air. 

Illustration:  A  converter  has  an  outside  surface  of  50  square 
meters,  at  an  average  temperature  during  the  blow  of  200°  C. 
the  average  air  current  being  1  meter  per  second,  and  out- 
side air  30°  C.  What  is  the  heat  loss  by  conduction  to  the 
air  in  kilogram  Calories  per  minute? 

Solution : 

Coefficient  of  transfer  conductivity : 

0.000028  (2+  VlOO)  =  0.000336  . 
Heat  loss  per  1°  difference,  per  second,  in  gram- calories : 

50X10,000X0.000336  =  168  calories. 

Heat 'loss  per  170°  difference,  per  minute,  in  kg.-calories: 
168  X 170  X  60 -T- 1000  =  1714  calories. 


THE  BESSEMER  PROCESS.  3t53 

Heat  Radiated  During  the  Blow. 

This  is  a  function  of  the  temperature  of  the  outside  shell,  the 
mean  temperature  of  the  surroundings  of  the  converter  and 
the  nature  of  the  metallic  surface.  As  the  surface  is  oxidized 
iron,  it  would  lose  about  0.0141  gram-calories  from  each  square 
centimeter  per  second,  if  at  a  temperature  of  100°  and  the 
surroundings  at  0°;  or  practically  1  gram-calorie  per  second 
from  each  square  meter,  for  every  100,000,000  of  numerical  dif- 
ference between  the  fourth  powers  of  the  absolute  tempera- 
ture of  the  radiating  surface  and  its  surroundings.  (See  Metal- 
lurgical Calculations,  Part  1.,  p.  185). 

Illustration :  Assuming  the  surroundings  of  the  converter  at 
30°,  in  the  preceding  illustration,  what  amount  of  heat  is  radiated 
per  minute  in  large  Calories? 

Solution:  The  absolute  temperatures  in  question  are  273  + 
30  =  303,  and  273  +  200  =  473.  The  difference  of  their  fourth 
powers  is : 

4734— 3034  =  41,626,500,000, 

which,  divided  by  100,000,000,  gives  416.265  gram-calories 
lost  per  second  per  each  square  meter  of  radiating  surface. 
The  radiation  loss  for  the  whole  surface  per  minute  is,  therefore: 

416.265  X  50  X  60  4- 1000  =  1249  kg.-Calories. 

While  the  assumption  made  as  to  the  temperature  of  the  out- 
side shell  is  doubtless  only  approximate,  yet  if  the  tempera- 
ture of  the  same  is  carefully  determined  the  radiation  loss  can 
be  accurately  calculated.  If  the  outside  of  the  converter  were 
polished,  this  radiation  loss  might  be  reduced  nearly  nine 
tenths. 

Problem  68. 

From  the  data  and  results  of  calculation  of  Problem  65  (see 
pages  310,  317  and  323,)  we  see  that  22,500  pounds  of  pig 
iron  and  2,500  pounds  of  spiegeleisen  produced  24,665  pounds 
of  steel,  there  being  eliminated  during  the  blow  and  recar- 
burization : 

Carbon 679.5    pounds  (140.7  to  CO2) 

Silicon 203.2 

Manganese 197.9 

Iron 253.9       "       (25.3  to  Fe2Os) 


364  METALLURGICAL  CALCULATIONS. 

The  gases  contain : 

CO2 5.20  per  cent. 

CO 19.91 

H2 1.39 

N2 73.50 

The  slag  contains: 

SiO2 ...63.56 

A12O3 3.01 

FeO 21.39 

Fe2O3 2.63 

MnO 8.88 

CeO 0.90 

MgO 0.36 

Make  average  assumptions  for  requisite  data  not  given. 

Required:     A  balance  sheet  of  heat  evolved  and  distributed. 

Solution:  The  items  of  this  balance  sheet  have  already  been 
discussed  in  detail.  We  will  apply  them  to  this  specific  case: 

Heat  in  Body  of  Converter  at  Starting:  Assuming  that  this 
is  a  blow  in  regular  running,  the  heat  may  be  taken  at  any 
reasonably  approximate  quantity,  because  the  same  quantity 
with  only  a  slight  deduction  will  be  allowed  as  contained  in 
the  same  on  finishing.  We  will,  therefore,  take  a  figure  already 
calculated,  8,034,970-pound  Calories,  as  the  heat  in  the  con- 
verter body  at  starting. 

Heat  in  Melted  Pig  Iron:  We  will  take  it  at  300  Calories 
per  pound,  or  a  total  of  300X22,500  =  6,750,000  Calories. 

Heat  in  Spiegeleisen:     2,500 X    300  =  750,000  Calories. 

Heat  in  Blast:  This  may  safely  be  considered  as  warmed 
by  compression  and  entering  the  converter  at  60°  C.  The 
amount  of  blast  received  altogether  is  calculated  thus: 

Carbon  oxidized  =     679.5  pounds. 

Volume  of  CO  and  CO2  formed: 

679.5X16^0.54  =  20,133  cu.  ft. 

Volume  of  gases  ^TT  =  80'179      " 

U.^Oll 

Volume  of  nitrogen  80,179  X  0.735    =  58,932      " 
Volume  of  air  proper  in  blast  =  74,408 

Volume  of  hydrogen  in  gases . 

80,179X0.0139=     1,115 
Volume  of  moisture  in  gases  =    1,115      " 


THE  BESSEMER  PROCESS.  365 

Assuming  the  blast  at  60°  C.,  the  heat  in  it  is: 

Air      74,408X0.3046    =     22,665  oz.  Cal.  per  1° 
H2O        1,115X0.3790  =          423      " 


23,088      - 

23,088X60  =  1,385,280  oz.  Cal. 

=  86,580  Ib.  Cal. 
Heat  of  Oxidation: 

C  to  CO2  140.7X8100  =  1,139,670  Calories 

C  to  CO  538.8X2430  =  1,309,280 

Si  to  SiO2  203.2X7000  =  1,422,400 

Mn  to  MnO  197.9X  1653  =     327,130 

Fe  to  FeO  228.6  X 1173  =     268,150 

Fe  to  Fe2O3  25.3X1746  =       44,170 


4,510,800 

Heat  of  Formation  of  Slag:  The  197.9  pounds  of  manganese 
oxidized  forms  255.5  of  MnO.  Hence  the  weight  of  the  slag 
is  255.5 -T- 0.0888  =  2877  pounds.  The  slag,  therefore,  con- 
tains also  2877X0.6356  =  1829  pounds  of  SiO2,  of  which  203.2 
X  60/28  =  435  pounds  came  from  the  silicon  oxidized,  and 
1794  pounds  from  the  lining.  The  lining  will  also  have  lost 
the  A1203,  CaO  and  MgO  in  the  slag,  equal  to  2877  X  0.0427 
=  123  pounds.  The  total  iron  going  into  the  slag,  253.9 
pounds,  is  equivalent  to  326.4  pounds  of  FeO. 

The  heat  of  formation  of  the  slag  will  therefore  be: 

FeO  326.4X124  =  40,474  Calories. 

MnO  255.5X   76  =  19,418 


Total  59,892 

Heat  in  Converter  at  Finishing:  This  will  be  the  same  as  at 
starting,  less  the  heat  in  1794  +  123  =  1917  pounds  of  lining, 
which  was  corroded  and  entered  the  slag.  Assuming  this  to 
have  been  on  the  inner  surface,  at  an  average  temperature  of 
1500°,  the  heat  in  it,  using  the  mean  specific  heat  of  silica,  will 
have  been 

1917X0.2988X1500  =  851,200  Calories. 


366  METALLURGICAL  CALCULATIONS. 

And  the  heat  in  the  converter  body  at  the  finish: 

8,034,970—851,200  =  7,183,770  pound  Calories 

Heat  in  Finished  Steel:     Taking  its  temperature  as   1650°, 
with  350  Calories  per  unit,  we  have 

24,665X350  =  8,632,750  Calories. 
Heat  in  the  Slag: 

2877X550  =  1,582,350 

Heat  in  Escaping  Gases:     These  have  already  been   calcu- 
lated as  consisting  of 

Nitrogen 58,932  cubic  feet. 

Hydrogen 1,115      "       " 

Carbon  monoxide 15,964      " 

Carbon  dioxide 4,169      " 

The  first  three  have  the  same  heat  capacity  per  cubic  foot,  so 
assuming  their  temperature  1550°: 

N2,  H2,  CO      76,011X534.5  =  40,627,900  oz.  Cal. 
CO2       4,169X947.1  =     3,948,250 


44,576,150 
=   2,786,000  Ib.  Cal. 

Absorbed  in  Decomposing  Moisture:  The  1,115  cubic  feet 
of  hydrogen  in  the  gases  represent  so  much  steam  or  water 
vapor  decomposed.  Since  1  cubic  foot  =  0.09  ounces,  the 
heat  absorbed  is 

1,115X0.09X29,040  =  2,914,160  oz.  Cal. 
=      182,130  Ib.  Cal. 

Heat  Conducted  to  the  Air:  Assuming  the  conditions  worked 
out  in  the  illustration  under  this  heading,  this  item  would 
be  approximately,  for  9  min.  10  sees.,  in  Ib.  Cal. 

1,714X2.204X9.167  =  34,630  Ib.  Cal. 

Heat   Lost   by   Radiation:     Making    similar    assumption,    we 

have 

1,249X2.204X9.167  =  25,240  Ib.  Cal. 


THE  BESSEMER  PROCESS.  367 

Recapitulation. 

Lb.  Cal. 

Heat  in  converter  body  at  starting 8,034,970 

melted  pig-iron 6,750,500 

spiegeleisen 750,000 

blast 86,580 

Heat  of  oxidation 4,510,800 

"        formation  of  slag 59,890 

Total  on  hand  and  developed 20,192,740 


Heat  in  converter  body  at  finish 7,183,770 

finished  steel 8,632,750 

slag 1,582,350 

escaping  gases 2,786,000 

Heat  absorbed  in  decomposing  moisture 182,130 

Heat  conducted  to  the  air 34,630 

Heat  lost  by  radiation 25,240 


Total  accounted  for 20,426,870 

Another  way  of  expressing  this  balance  is  to  itemize  the 
avenues  of  heat  evolution  and  utilization,  as  follows: 

Lb.  Cal. 

Received  from  converter  body 851,200 

Received  by  oxidation 4,510,800 

Received  by  formation  of  slag 59,890 

Total 5,421,890 

Lb.  Cal. 

Used,  excess  of  heat  in  gases  over  blast 2,699,420 

Used,  excess  of  heat  in  steel  and  slag  over  pig  iron  and 

spiegel 2,714,600 

Decomposition  of  moisture 182,130 

Radiation  and  conduction .       59,870 


Total 5,656,020 


CHAPTER    IX. 

THE  TEMPERATURE  INCREMENT  IN  THE  BESSEMER 

CONVERTER. 

In  the  preceding  chapter,  we  have  studied  the  generation  of 
heat  in  the  Bessemer  converter,  arid  its  distribution.  We  saw 
in  that  analysis  that  in  a  typical  operation,  nearly  one-half  of 
the  heat  generated  during  the  blow  is  carried  out  by  the  hot 
gases,  about  half  is  represented  in  the  increased  temperature  of 
the  contents  of  the  converter,  while  only  about  5  per  cent,  is 
lost  by  radiation,  etc.  In  the  present  paper  we  wish  to  analyze 
still  further  this  question  of  increased  temperature,  which  is  so 
vitally  necessary  for  the  proper  working  of  the  process,  and  to 
calculate  the  relative  efficiency  of  the  various  substances  oxidized 
in  causing  this  rise  of  temperature. 

While  at  no  time  in  the  Bessemer  operation  is  only  one  sub- 
stance being  oxidized,  yet  we  can  get  the  best  basis  for  our 
computations  by  assuming  a  charge  in  operation  and  only  one 
substance  oxidized  at  a  time.  Whatever  substance  is  in  ques- 
tion, let  us  assume  one  kilogram  burnt  in  a  given  short  period 
of  time,  generating  the  heat  of  its  combustion.  With  the 
bath  at  a  given  temperature,  the  air,  at  say  100°  C.,  comes 
in  contact  with  it,  bearing  the  necessary  oxygen.  If  nothing 
was  oxidized,  the  oxygen  and  nitrogen  would  simply  be  heated 
to  the  temperature  of  the  bath,  and"  pass  on  and  out,  while  the 
bath  would  be  meanwhile  losing  heat  also  by  radiation.  It 
is  evident  then,  that  unless  at  least  as  much  heat  as  the  sum 
of  these  two  items  is  generated,  the  bath  will  cool,  and  that 
only  the  excess  of  heat  above  this  requirement  is  available  for 
increasing  the  temperature  ot  the  bath  and  resulting  gases. 
The  proper  procedure  for  us  will  therefore  be  to  calculate, 
in  each  case,  the  chilling  effect  of  the  air  entering,  subtract 
this  from  the  heat  generated,  and  the  residue  is  net  heat  avail- 
able for  raising  the  temperature  of  the  contents  of  the  con- 

368 


THE  BESSEMER  CONVERTER.  369 

verier  and  the  gases,  and  supplying  radiation  losses.  The 
latter  are  proportional  to  the  time,  and  therefore  to  the  amount 
of  air  used,  assuming  blast  constant. 

SILICON. 

This  is  burnt  out  in  the  first  part  of  the  process,  during  which 
the  temperature  begins  low  and  ends  high.  We  will  there- 
fore assume  two  temperatures,  and  calculate  the  thermal 
increment  at  each.  We  will  take  1250°  and  1600°.  The  net 
heat  is  absorbed  by  the  bath,  slag  and  nitrogen. 

Oxygen  necessary  to  burn  one  kilogram  of  silicon : 

IX 32^28=  1.143    kg. 

Nitrogen  accompanying  this  oxygen  =  3.810 

Weight  of  air  needed  =  4.953      " 

Volume  of  air  =  4.953-^-1.293  =  3.831    m3 

Specific  heat,  100°  to  1250°,  per  m3  =  0.3395 

Specific  heat,  100°  to  1600°,  per  m3  =  0.3489 

Chilling  effect  of  air  at  100°,  bath  at  1250° 

=  3.831X0.3395X1150  =      1496  Cal. 

Shilling  effect  of  air  at  100°,  bath  at  1600° 

=  3.831  X  0.3489  X 1500  =      2043  Cal. 

Heat  generated  per  kilogram  of  silicon       =     7000  Cal. 

This  is,  however,  for  cold  oxygen  and  cold  silicon  burning 
to  cold  solid  silica.  Under  the  conditions  prevailing  we  have 
melted  silicon  at  the  temperature  of  the  bath,  oxidized  by  hot 
oxygen  to  hot  silica,  giving  a  slightly  different  heat  of  com- 
bination, calculated  as  follows: 

Heat  in  melted  silicon  at  1250°  =  480  Cal. 

"     "      oxygen  required,  at  1250°  =  334     " 

"     "      silica,  at  1250°  =  750     " 

Heat  of  oxidation  at  1250° 

=  7,000  +  480  +  334—750  =  7,064    " 

The  difference  from  7,000  is  so  small  as  to  be  within  the 
possible  error  of  the  7,000  itself,  and  we  can  therefore  make 
the  calculations  with  all  the  accuracy  they  allow,  taking  the 
ordinary  heats  of  oxidation  from  the  tables. 

One  factor  of  heat  generation  has,  however,  not  been  men- 
tioned, viz.:  the  heat  of  combination  of  silica  with  oxides  of 


370  METALLURGICAL  CALCULATIONS. 

iron  and  manganese  to  form  slag.  This  is  148  Calories  per  kg. 
of  silica  when  forming  iron  silicate,  and  90  when  forming  man- 
ganese silicate,  which  quantities  would  become  317  and  193 
Calories  respectively  when  calculated  per  kg.  of  silicon  oxi- 
dizing. The  question  arises,  however,  whether  it  is  fair  to 
credit  all  this  to  the  silica,  because  this  generation  of  heat  by 
slag  formation  is  really  a  mutual  affair,  chargeable  to  the  credit 
of  both  silica  and  the  other  oxides;  we  should,  therefore,  not 
charge  it  all  up  to  the  credit  of  silica  formation,  and  we  do 
not  know  what  part  to  charge  to  the  credit  of  silica  if  we  do 
not  charge  it  all.  In  this  dilemma  it  may  be  well  to  remem- 
ber that  silicon  is  probably  oxidized  before  the  iron  and  man- 
ganese, and  that  the  heat  of  formation  of  the  slag  is  therefore 
more  properly  considered  as  being  generated  afterwards,  and 
therefore  may  be  practically  credited  entirely  to  the  oxidation 
of  iron  and  manganese. 

Resume  for  oxidation  of  1  kilo,  of  silicon: 

Cal. 

Heat  generated  =  7,000 

Chilling  effect  of  the  blast,  100°  to  1250°  =  1,496 

Chilling  effect  of  the  blast,  100°  to  1600°  =  2,043 

Available  heat,  bath  at  1250°  =  5,504 

Available  heat,  bath  at  1600°  =  4,957 

If  we  assume  a  radiation  loss  proportional  to  the  length  of 
the  blow,  i.  e.,  proportional  to  the  air  blown  in,  we  can  find 
out  from  average  blows  that  this  amounts  to  about  50  Calories 
per  cubic  meter  of  blast  used.  The  radiation  loss  during  com- 
bustion of  1  kilogram  of  silicon  would  therefore  be 

Cal. 

3.831X50  =      192 

Leaving  net  available  heat  at  1250°  =  5,312 

At  1600°  =  4,765 

The  above  quantity  of  heat  is  expended  in  raising  the  tem- 
perature of  99  kg.  of  bath,  2.143  kg.  of  silica,  and  3.810  kg.  of 
nitrogen  gas,  from  their  initial  temperature.  At  the  tempera- 
tures of  1250°  and  1600°,  respectively,  the  heat  capacity  of 
these  products  of  the  operation  will  be,  per  1°  C.  rise: 


THE  BESSEMER  CONVERTER.  371 

Specific  Heat  Heat  Capacity 

Products.  at  1250°     at  1600°  at  1250°     at  1600° 

Bath,  99  kg 0  25        0.25  24.8         24.8 

SiO2,  2.14  kg 0.37         0.43  0.8  0.9 

N2,  3.02m3..  0.37         0.39  1.1  1.2 


Totals.         26.7  26.9 

Theoretical  rise  of  temperature : 

5,312 -v- 26.7  =  199°  (bath  at  1250°) 

4,765^26.9  =  177°  (bath  at  1600°) 

Average  rise,  per  average 

1%  of -silicon  =188°  C. 

MANGANESE. 

As  high  as  4  per  cent,  of  manganese  may  be  oxidized  during 
the  blow,  and  therefore  this  heat  of  combustion  is  sometimes 
important.  We  will  calculate  the  net  heat  available  for  rais- 
ing temperature  and  the  rise  of  temperature  per  1  per  cent,  of 
manganese  oxdized,  i.  e.,  for  1  kilo,  of  manganese  per  100 
kilos,  of  bath. 

Oxygen  necessary  1 X 16/55  =  0.291  kg. 

Nitrogen  accompanying  this  =  0.970 

Weight  of  air  used  =  1.261      " 

Volume  of  air  used  =  0.975  m3 

Chilling  effect  of  air  at   100°  on  bath  at 

1250°  =  0.975X0.3395X1150  =  381  Cal. 

Heat  generated  per  kg.  of  manganese          =  1653     " 
Heat  of  formation  of  MnO.SiO2 

1.291kg.  MnOX  76  =  98    " 

Total  heat  developed  =  1751     " 

Heat  available  =  1751—381  =  1370    " 

Radiation  loss  =  0.975X50  =  49    " 

Net  available  heat  =  1321     " 

Heat  capacity  of  99  kg.  of  bath  per  1°        =  24.8    " 

Heat  capacity  of  2.4  kg.  of  slag  =  0.7    " 

Heat  capacity  of  0.8  m3  nitrogen  =  0.3     " 

Heat  capacity  of  products  per  1°  =  25.8    " 


372  METALLURGICAL  CALCULATIONS. 

Theoretical  rise  of  temperature : 

1321^25.8  =  51°  C. 
This  is,  in  round  numbers,  one-fourth  the  efficiency  of  silicon. 

IRON. 

While  it  is  not  desired  to  oxidize  iron,  and  while  it  is  rela- 
tively less  oxidizable  than  silicon  or  manganese,  yet  some 
is  always  oxidized  because  of  the  great  excess  of  iron  present 
in  the  converter.  The  amount  of  iron  thus  lost  is  variable,  and 
some  of  it,  towards  the  end  of  the  blow,  may  be  oxidized  to 
Fe203  instead  of  FeO,  the  larger  part,  however,  oxidizes  to 
FeO.  We  will  make  the  calculations  for  both  oxides,  per  kilo- 
gram of  iron. 

Formation  of  FeO. 

Oxygen  necessary  1X16/56  =  0.286  kg. 

Nitrogen  accompanying  this  =  0.953    " 

Weight  of  air  used  =  1.239    " 

Volume  of  air  used  =  0.958  m3 

Chilling  effect  of  air  at  100°  on  bath  at 

1250°  =  0.958X0.3395X1150  =      374  Cal. 

Chilling  effect  of  air  at  100°  on  bath  at 

1600°  =  0.958  X. 3489X1500  =     501    " 

Heat  generated  per  kg.  of  iron  =1,173    " 

Heat  of  formation  of  FeO.SiO2  =  1.286kg. 

FeO  X 124  =      159    " 

Total  heat  developed  =  1,332    " 

Net  heat  available  at  1250°  =  1332— 374  =  958  " 
Net  heat  available  at  1600°  =  1332— 501  =  821  " 
Radiation  losses  =  0.958X50  =  48  " 

Net  available  heat  at  1250°  =     910    " 

Net  available  heat  at  1600°  =     773    " 

Heat  capacity  of  99  kg.  of  bath  per  1°  =  24.8  " 
Heat  capacity  of  2.4  kg.  of  slag  per  1°  =  0.7  " 
Heat  capacity  of  0.8  m3  of  nitrogen  =  0.3  " 

Heat  capacity  of  products  per  1°  =    25.8    * 

Theoretical  rise  of  temperature: 

910^25.8  =  36°  C. 
773  H-  25.8  =  30°  C. 

This  is  only  about  one-sixth  as  efficient  as  silicon. 


THE  BESSEMER  CONVERTER.  373 

Formation  of  Fe20*. 

Weight  of  air  used  =  1.859  kg. 

Volume  of  air  used  =  1.438  m3 
Chilling  effect  of  air  at  100°  on  bath  at 

1600°  =  1.438X0.3489X1500  =     753  Cal. 

Total  heat  developed  by  oxidation  =    1746    " 

Heat  of  formation  of  slag  =      159    " 

Total  heat  developed  =    1905    " 

Heat  available  =  1905—753  =    1152    " 

Radiation  losses  =  1.438X50  =        72    " 

Net  heat  available  =    1080    " 

Heat  capacity  of  products  =       26    " 
Theoretical  rise  of  temperature : 

1080^26  =  42°  C. 

TITANIUM. 

While  titanium  is  an  unusual  constituent  of  pig  iron,  yet  it  is 
conceivable  that  titaniferous  pig  iron  might  be  made  and  blown 
to  steel.  If  so,  the  following  calculation,  based  on  a  quite  re- 
cently determined  value  for  the  heat  of  oxidation  of  titanium, 
will  show  that  the  titanium  is  a  valuable  heat  producing  sub- 
stance, being,  in  fact,  weight  for  weight  three  fourths  as  efficient 
as  silicon. 

Oxygen  needed  1x32/48  =  0.667  kg. 

Nitrogen  accompanying  this  =  2.222    " 

Air  used  =  2.888   " 

Volume  of  air  needed  =  2.250  m* 
Chilling  effect  of  air  at  100°  on  bath  at 

1250°  =  2.250X0.3395X1150  =     878  Cal. 

Heat  generated  per  kg.  of  Ti  =   4542    " 

Heat  of  formation  of  slag — unknown 
Net  heat  available  =  4542—878  =    3664    " 

Deducting  144  for  radiation  losses  =   3520    " 

Heat  capacity  of  products  per  1°  C.  =    26.5    " 

Theoretical  rise  of  temperature : 

3520-5-26.5  =  133°  C. 


374  METALLURGICAL  CALCULATIONS. 

ALUMINIUM. 

This  metal  is  also  rarely  found  in  pig  iron,  yet  when  present 
it  would  be  a  powerful  heat  producer,  as  the  following  calcu- 
lations show: 

Oxygen  needed  1x48/54  =  0.889  kg. 

Nitrogen  =  2.963    " 

Air  =3.852    " 

Volume  of  air  =  2.964  m3 

Chilling  effect  of  air  at  100°  on  bath  at 

1250°  =  2.964X0.3395X1150  =    1157  Cal. 

Heat  generated  per  kg.  of  Al  =    7272    " 
Heat  of  formation  of  slag — uncertain 

Heat  available  =  7272—1157  =    6115    " 

Deducting  for  radiation  losses  =    5967    " 

Calorific  capacity  of  products  per  1°  C.  =    26.6    " 
Theoretical  rise  of  temperature : 

5967 -r  26.6  =  224°  C. 

If  a  blow  was  running  cold,  and  ferro-silicon  was  not  on 
hand  to  add  in  order  to  increase  its  temperature,  ferro-alum- 
inium,  or  aluminium  itself,  would  be  a  good  substitute  in  the 
emergency. 

NICKEL. 

It  is  hardly  probable  that  nickeliferous  pig  iron  would  be 
blown  to  steel,  because  of  the  waste  of  valuable  nickel  in  the 
slag;  yet  if  1  per  cent,  of  nickel  were  thus  oxidized,  calculations 
similar  to  the  preceding  would  show  a  net  rise  in  temperature 
of  the  contents  of  the  bath  of  about  33°  C. 

CHROMIUM. 

Quite  recently  some  chromiferous  pig  iron  has  been  blown 
in  the  Bessemer  converter,  and  those  in  charge  were  hampered 
by  the  lack  of  data  as  to  how  chromium  would  behave  during 
the  blow  and  its  heat  value  to  the  converter,  Technical  litera- 
ture will  probably  soon  contain  an  account  of  the  practice 
which  has  been  developed  at  the  Sparrow's  Point  works  of  the 
Maryland  Steel  Company.  No  thermo-chemist  has,  as  yet, 
determined  the  heat  of  formation  of  chromium  slag.  From 


THE  BESSEMER  CONVERTER.  375 

what  we  know  of  the  chemical  reactions  of  chromium  it  is 
likely  that  this  heat  is  considerable.  Neglecting  the  heat  of 
formation  of  slag,  we  have  the  following  approximation  to  its 
heating  efficiency,  assuming  it  to  be  oxidized  when  the  bath 
is  near  to  its  maximum  temperature: 

Oxygen  needed  1x48/104  =  0.462  kg. 

Nitrogen  entering  =  1.540    " 

Air  used  =  2.002    " 

Volume  of  air  =  1.548  tn3 
Chilling  effect  of  air  at  100°  on  bath  at 

1600°  =  1.548X0.3489X1500  =      810  Cal. 

Heat  of  oxidation  =  2,344    " 

Net  heat  =  2344—810  =  1,534    " 

Deducting  for  radiation  losses  =  1,457    " 

Calorific  capacity  of  products  per  1°  =    26.1    " 

Theoretical  rise  of  temperature : 

1457-26.1  =56°C. 

CARBON. 

This  element  commences  to  be  oxidized  in  large  amount  only 
towards  the  middle  of  the  blow,  when  the  temperature  of  the 
bath  is  high,  because  of  the  previous  oxidation  of  silicon.  It 
will  be  about  right,  therefore,  to  estimate  the  bath  at  an 
average  temperature  of  1600°  during  the  elimination  of  carbon. 
The  product  is  mostly  CO,  but  partly  CO2.  We  will,  therefore, 
calculate  for  each  of  these  possible  products  separately.  The 
net  heat  available,  after  allowing  for  average  radiation  losses, 
is  used  to  increase  the  temperature  of  the  products,  i.  e.,  of  the 
bath,  the  nitrogen  and  the  CO  or  CO2. 

Oxidation  to  CO2. 

Oxygen  required  =1x32/12  =  2.667  kg. 

Nitrogen  accompanying  =  8.889    " 

Air  used  =  11.556    " 

Volume  of  air  =  8.937  m3 

Chilling  effect  of  air  at  100°  on  bath  at 

1250°  =  8,937X0.3395X1150  =  3,489  Cal. 

Heat  of  oxidation  =  8,100    " 


376  METALLURGICAL  CALCULATIONS. 

Heat  available  =  8,100—3,489                   =  4,611  Cal. 

Radiation  losses  =  8.937X50                      =  447  " 

Net  heat  available                                        =  4,164  " 

Heat  capacity  of  99  kg.  bath  =  99X0.25=  24.80  " 

Heat  capacity  of  7.05  m3  N2  =  7.05X0.37  =  2.61  " 

Heat  capacity  of  1.90m3  CO2  =  1.90X0.88=  1.70  " 

Heat  capacity  of  products,  per  1°  C.          =  29.11  " 

Theoretical  rise  of  temperature : 

4,164-*- 29.11  =143°C. 

Since  carbon  burns  to  CO2  principally  at  the  beginning  of 
the  blow,  while  the  bath  is  cold,  we  see  that  carbon  thus  con- 
sumed is  about  three-quarters  as  efficient  as  an  equal  weight  of 
silicon  in  raising  the  temperature  of  the  bath. 

Oxidation  to  CO. 

Oxygen  needed  1 X 16/12  =  1.333  kg. 

Nitrogen  accompanying  =  4.444    " 

Air  used  =  5.777    " 

Volume  of  air  =  4.469  m3 
Chilling  effect  of  air  at  100°  on  bath  at 

1600°  =  4.469X0.3489X1500  =  2,339  Cal. 

Heat  of  oxidation  =  2,430    " 
Heat  available  =  2,430—2,339  91    " 

Radiation  losses  =  4.469X50  =      233    " 

Net  heat  available  =  91—233  =—142     " 

Heat  capacity  of  99  kg.  of  bath  =  99X0.25  =     24.8    " 

Heat  capacity  of  3.5  m3  of  N2  \  ,.  .     ft  oQ  _        9  *     „ 
Heat  capacity  of  1.9  m3  of  CO  /  M*™fl 

Heat  capacity  of  products  =    26.9    " 

Theoretical  rise  of  temperature : 

-142 -r- 26.9  =  — 5°C. 

The  result  is,  therefore,  that  when  carbon  burns,  as  it  mostly 
does,  to  CO,  and  the  temperature  of  the  bath  is  high,  there  is 
practically  no  further  rise  of  temperature,  for  the  heat  of  oxi- 
dation is  barely  sufficient  to  counteract  the  chilling  effect 
of  the  air  and  to  supply  radiation  and  conduction  losses. 

In  the  above  calculations,  no  allowance  was  made  for  heat 


THE  BESSEMER  CONVERTER.  377 

required  to  separate  carbon  from  its  combination  with  iron, 
or  for  the  variation  in  the  heat  of  combination  of  carbon  with 
oxygen  from  the  combination  heats  at  ordinary  temperatures. 
The  former  is  not  known,  or  perhaps  is  very  nearly  zero.  The 
heat  of  oxidation  of  liquid  carbon  at  1250°  to  CO2  or  at  1600° 
to  CO  is  calculated  as  follows: 

Oxidation  1  kg.  C  to  CO2  at  0°  =  8,100  Cal 

Heat  to  raise  1  kg.  C  to  1250°         =  505  Cal. 
Heat  to  liquefy  1  kg.  at  1250°         =  129    " 
Heat  to  raise  2.67  kg.  O2  to  1250°  =  779    " 

Heat  to  raise  reacting  substance  to  1250°  =  1,413 

Heat  in  3.67  kg.  CO2  at  1250°  =  1,493    " 

Heat  of  reaction  at  1250°  (8,100+1,413-1,493)       =  8,020    " 

For  production  of  CO,  at  1600°,  the  correction  is  larger, 
as  is  seen  from  the  following : 

Oxidation  1  kg.  C  to  CO  at  0°  =  2,430  Cal. 

Heat  to  raise  1  kg.  to  1600°  =  680  Cal. 

Heat  to  liquefy  1  kg.  C  at  1600°     =  156    " 
Heat  to  raise  1.33  kg.  O2  to  1600°      =  554    " 

Heat  to  raise  reacting  substances  to  1600°  =  1,390    " 

Heat  in  2.33  kg.  of  CO  at  1600°  =  1,104    " 

Heat  of  reaction  at  1600°  (2,430  +  1,390—1,104)         =  2,716    " 

The  use  of  this  corrected  value  makes  the  oxidation  of  C 
to  CO  give  a  small  net  heat  development,  with  consequent 
slight  rise  of  temperature,  instead  of  the  slight  cooling  effect 
before  calculated.  The  conditions  are  so  nearly  even,  however, 
that  a  slight  increase  of  temperature  or  slowing  up  of  the  blow 
would  wipe  out  the  heat  excess. 

PHOSPHORUS. 

This  is  the  last  important  element  to  be  considered,  and  is 
always  eliminated  after  the  carbon,  at  the  maximum  bath  tem- 
perature, which  we  will  assume,  for  calculation,  at  1600°.  The 
heat  generated,  at  ordinary  temperatures,  is  5892  Calories  per 
kilogram  of  solid  phosphorus.  Per  kilogram  of  liquid  phos- 
phorus it  would  be  only  5  Calories  more,  or  5897  Calories.  For 
the  reaction  at  1600°  we  would  have  a  different  value,  probably 


378  METALLURGICAL  CALCULATIONS. 

some  500  Calories  more,  but  the  necessary  data  concerning  the 
specific  heats  of  P  and  P2O5  are  not  known,  and  we  must  omit 
this  calculation.  The  heat  of  combination  of  iron  and  phos- 
phorus is  also  a  doubtful  quantity.  Ponthiere  places  it  as  high 
as  1,397  Calories  per  kilogram  of  phosphorus,  but  this  ap- 
pears altogether  improbable  since  another  experimenter  could 
obtain  no  heat  of  combination  at  all.  As  concluded  in  another 
place,  I  advise  for  the  present  omitting  this  questionable 
quantity. 

The  phosphorus  pent-oxide  forms  3CaO .  P2O5  with  the  lime 
added,  but  since  there  is  always  more  lime  present  than  cor- 
responds to  these  proportions  (3CaO  :  P205  ::  168  :  142),  the 
calculation  of  heat  of  formation  of  the  slag  must  be  based  on 
the  amount  of  P2O5  formed  (1123  Calories  per  kilogram  of 
P205).  This  amounts  to  a  considerable  item.  On  the  other 
hand,  the  lime  needed  for  slag  is  put  in,  usually  preheated,  for 
the  sole  purpose  of  combining  with  the  P2O5.  It  seems,  there- 
fore, only  right  to  charge  the  phosphorus  with  the  heat  re- 
quired to  raise  this  lime  to  the  temperature  of  the  bath.  The 
lime  added  averages  three  times  the  weight  of  P2O5  formed, 
and  is  preheated  usually  to  about  600°.  Assuming  these  con- 
ditions, the  following  calculations  can  be  made  per  kilogram 
of  phosphorus  oxidized : 

Oxygen  required  =  1.29  kg. 

Nitrogen  accompanying  =  4.30 

Air  used  =  5.59    " 

Volume  of  air  =  4.32m3 
Heat  of  formation  of  slag, 

2.29  kg.  P2O5X1123  =  2572  Cal. 

Heat  of  oxidation  of  phosphorus  =  5897 

Total  heat  developed  =  8469    " 

Chilling  effect  of  air  at  100°  on  bath  at 

1600°  =  4.32X0.3489X1500  =  2261 

Chilling    effect    of   lime    (600°  to    1600°) 

(2.29 X  3)  X 0.328X1000  =  2253 

Chilling  effect  of  blast  and  lime  =  4514 

Heat  available  =  8469—4514  =  3955 

Radiation  losses  =  4.32X50  =      216 

Net  heat  available  =  3739    " 


THE  BESSEMER  CONVERTER.  379 

Heat  capacity  99  kg.  of  bath  =  99  X  0.25      =     24.8  kg. 
Heat  capacity  3.4m3  of  N2  =  3.4X0.39        =       1.3    " 
Heat  capacity  6.9  kg.  of  slag  =  6.9X0.3       =       2.1     " 

Heat  capacity  of  products,  per  1°  =     28.2    " 

Theoretical  rise  of  temperature  : 

3739^-28.2  =  133°  C. 

If  the  lime  were  added  cold,  its  cooling  effect  would  be  883 
Calories  greater,  the  net  heat  available  would  be  883  Calories 
less,  and  the  calculated  rise  of  temperature  31°  less,  or  102°  C. 
Using  the  preheated  lime  we  can  regard  phosphorus  as  being 
practically  two-thirds  as  efficient,  weight  for  weight,  as  silicon; 
with  cold  lime,  about  one-half  as  efficient. 


RESUME. 
Heat  effect  of  oxidizing  1  kilogram  of  element. 


or 


f 
u 


J 

^ 

a 

|| 

a 
•& 

1 

1 

aj 

tfc^ 
^^ 

^ 

+4 
J 

1 

'o 

•$ 

11 

* 

fc, 

(O  Qq 

Silicon               

7  000 

7  000 

1  688 

Manganese           ...    , 

.    .    .    .          1  653 

98 

1  751 

430 

Iron  (to  FeO)  

1,173 

159 

1,332 

422 

Iron  (to  Fe2O3)  .  . 

1,746 

159 

1  905 

825 

Titanium   

.  .4,542 

4,542 

1  022 

Aluminium               .  .  . 

7  272 

7  272 

1  305 

Nickel 

1  051 

159 

1  210 

378 

Chromium 

2  344 

2  344 

887 

Carbon  (to  CO2)     .  .  . 

8.100 

8  100 

3  936 

Carbon  (to  CO) 

2  430 

2  430 

2  572 

Phosphorus   

5  897 

2  572 

8  469  ( 

2  477 

2,253* 

*  Chilling  effect  of 

lime  added,  preheated 

to  600° 

a      »'lt.* 

It?  I 

!«•  il 

sSf  If 

^     O        -Si     1> 


5,312  188° 

1,321  51° 

910  33° 

1,080  42° 

3,520  133° 

5,967  224° 

832  33° 

1,457  56° 

4,164  143° 

-142  —5° 

3,739  133° 


It  must  be  observed  that  the  above  table  is  for  comparison 
only,  it  cannot  be  used  for  an  actual  case,  such  as  when  1  per 
cent,  of  silicon,  3  of  iron,  4  of  carbon  and  2  of  phosphorus  are 


380  METALLURGICAL  CALCULATIONS. 

oxidized.     In  such  a  case,  the  rise  in  temperature  would  be 
only  very  roughly: 

-  From  silicon 1 X 188  =  188° 

From  iron 3  X   33  =    99° 

From  carbon -     say  =      0° 

From  phosphorus 2X 133  =  266° 

Total =  553°  C. 

It  is  to  be  recommended  that  in  each  specific  case  the  calcu- 
lation be  made  for  the  specific  conditions  obtaining,  such  as 
temperature  of  the  metal  at  starting,  temperature  of  the  blast, 
time  of  the  blow  (as  far  as  this  affects  radiation  and  conduc- 
tion losses),  proportion  of  carbon  burned  to  CO2,  free  oxygen 
in  the  gases,  moisture  in  the  blast,  temperature  and  quantity 
of  lime  added,  corrosion  of  lining.  When  all  these  items  and 
conditions  are  taken  into  account  there  will  be  room  for  only 
small  discrepancy  between  the  calculated  and  the  observed  rise 
of  temperature.  The  chief  items  needing  experimental  re- 
search at  present  are:  The  specific  heat  of  the  melted  bath, 
the  specific  heat  of  the  slag,  the  heat  of  combination  of  various 
elements  comprising  the  bath,  the  heat  of  formation  of  the 
slag,  and  the  heat  of  oxidation  of  some  of  the  rarer  elements. 
Such  establishments  as  the  Carnegie  Institution  could  not  do 
the  cause  of  metallurgy  better  service  than  to  subsidize  metallur- 
gical laboratories  for  the  determination  of  such  data. 


CHAPTER  X. 
THE  OPEN-HEARTH  FURNACE. 

By  the  above  title  we  mean  to  designate  not  only  the  re- 
generative gas  furnaces  for  making  steel  but  also  those  for 
reheating  purposes;  in  other  words,  regenerative  or  recupera- 
tive reverberatory  furnaces.  Prominent  among  these  is  the 
Siemens-Martin  furnace,  with  complete  gas  and  air  preheating 
regenerative  chambers.  In  all  these  furnaces  the  charge  is 
heated,  or  kept  hot,  partly  by  direct  contact  with  the  gaseous 
products  of  combustion  and  partly  by  radiation  from  the  flame 
and  the  sides  and  roof  of  the  furnace.  It  was  the  Siemens 
brothers  who  first  insisted  on  the  relatively  great  importance 
of  the  radiation  principle,  in  distinction  to  the  direct  impinge- 
ment of  the  flame  on  the  material,  pointing  out  that  a  luminous 
flame  radiates  from  all  parts  of  its  volume,  while  a  hot,  solid 
body  radiates  only  from  its  surface,  and  direct  impingement 
interferes  with  the  development  of  perfect  combustion  and 
communicates  heat  relatively  slowly  at  best. 

GAS  PRODUCERS. 

Gas  producers  are  the  usual  adjunct  for  open-hearth  fur- 
naces, excepting  where  natural  gas  or  blast  furnace  gas  is 
available.  They  may  be  placed  far  from  the  furnace,  when 
they  deliver  cool  gas  to  the  regenerators,  or  close  to  the  fur- 
nace, delivering  comparatively  hot  gas  to  the  regenerators,  or 
even  be  made  part  of  the  furnace  itself,  delivering  their  hot 
gas  immediately  to  the  ports  of  the  furnace.  The  latter  is  un- 
doubtedly the  most  economical  arrangement  where  practicable. 

The  following  generalizations  concerning  the  relations  of  the 
gas  producer  to  the  open-hearth  furnace  may  be  made:  Pro- 
ducers furnish  4,300  to  4,600  cubic  meters  of  gas  per  metric 
ton  of  coal  used  (150,000  to  160,000  cubic  feet  per  short  ton) ; 
the  gas  produced  runs  3  to  8  per  cent.  CO2,  5  to  20  per  cent.  H2. 
20  to  30  per  cent.  CO,  and  50  to  60  per  cent.  N2;  its  calorific 

381 


382  METALLURGICAL  CALCULATIONS. 

power  is  750  to  1000  Calories  per  cubic  meter  (47  to  63-pound 
Calories,  or  85  to  115  B.  T.  U.  per  cubic  foot);  its  calorific 
power  represents  60  to  90  per  cent,  of  the  calorific  power  of  the 
fuel  used;  in  steel-making  processes,  the  keeping  of  the  furnace 
up  to  proper  heat  requires  the  gasifying  of  25  to  35  kilograms 
(50  to  80  pounds)  of  coal-  per  hour,  in  the  producers,  for  each 
ton  of  metal  capacity  of  the  furnace;  good  producers  gasify 
60  to  65  kilograms  of  coal  per  hour  per  each  square  meter  of 
gas-producing  area  (10  to  15  pounds  per  hour  per  each  square 
foot);  a  furnace  therefore  requires  some  0.4  to  0.6  square  meter 
(4  to  6.5  square  feet)  of  gas-producing  air  in  the  producers 
for  each  ton  of  metal  capacity  of  the  furnace. 

FLUES  TO  FURNACE. 

In  conducting  the  gases  to  the  furnace,  the  flues  or  conduits 
should  be  of  ample  size.  If  too  small  the  gas  must  pass  through 
them  with  high  velocity,  requiring  considerable  draft  to  give 
them  this  velocity,  which  the  chimney,  or  blower  may  or  may 
not  be  capable  of  furnishing.  Producers  are  almost  always 
worked  by  a  steam  blower,  furnishing  mixed  air  and  steam 
and  a  plenum  of  pressure  in  the  upper  part  of  the  producer, 
which  suffices  to  send  the  gas  through  the  conduits  under  a 
slight  pressure,  and  thus  avoids  any  sucking  in  of  air  through 
crevices  in  the  conduits.  With  too  sm'all  conduits  the  result- 
ing friction  and  high  velocity  required  may  give  the  blower 
more  work  than  it  can  do,  and  thus  entail  demands  for  draft 
upon  the  furnace  stack.  A  reasonable  rule  is  to  give  the  flues 
such  cross-sectional  area  that  the  hot  gas  which  must  pass 
through  them  shall  have  a  velocity  between  2  and  3  meters  per 
second. 

REGENERATORS. 

The  dimensions  of  the  regenerators  are  of  the  first  import- 
ance to  the  working  of  the  furnace.  They  should  have  suf- 
ficient length  in  the  direction  the  gas  currents  are  passing, 
so  that  the  gases  may  be  properly  cooled  or  heated ;  they  should 
have  sufficient  cross-sectional  area  of  free  space,  so  that  the 
velocity  of  the  gases  through  them  is  not  too  great;  they  must 
have  sufficient  thermal  capacity,  so  that  they  can  absorb  the 
requisite  quantity  of  heat. 

Length. — From  4  to  6  meters  (13  to  20  feet)  is  a  suitable 


THE  OPEN-HEARTH  FURNACE.  383 

length  in  the  direction  of  the  gas  currents.  This  permits  the 
hot  products  to  become  properly  cooled  before  going  to  the 
chimney,  and  the  gas  or  air  to  be  properly  heated  before  en- 
tering the  furnace.  The  shorter  length  may  be  used  when  the 
regenerator  is  of  large  cross-sectional  area,  with  slow  velocity 
of  gas  currents  through  the  free  spaces;  the  longer  when  the 
regenerator  is  rather  restricted  in  cross-section  and  the  gas 
currents  have  somewhat  high  velocity. 

Cross-Section. — The  free-space  sectional  area  should  be  such 
that  the  gases  where  hottest  should  not  have  a  calculated 
velocity  of  over  3  meters  (10  feet)  per  second,  and  if  calculated 
for  2  meters  (6.5  feet)  will  give  much  better  results  as  regards 
transmission  of  heat  to  the  checker  work.  In  this  manner, 
knowing  how  much  gas  must  go  through  the  regenerator  and 
what  its  maximum  temperature  will  probably  be,  the  cross- 
section  area  of  free  passage  space  can  be  calculated.  The  rela- 
tion of  this  to  the  cross-section  of  the  entire  stove  must  next 
be  considered,  and  this  is  entirely  a  question  of  how  the  checker 
work  is  built  up.  If  the  bricks  are  stacked  close  together  the 
free  space  may  be  reduced  to  as  much  as  one-half  the  total; 
as  ordinarily  stacked  it  may  be  60  to  80  per  cent,  of  the  total; 
if  perforated  bricks  are  used,  as  in  blast  furnace  firebrick- 
stoves,  the  area  of  free  space  averages  one-half  the  total ;  with 
ordinary  bricks  the  average  is  70  per  cent.  This  is  a  question 
which  is  very  variously  worked  out  in  different  furnaces,  and 
to  which  not  as  much  scientific  thought  has  been  given  as  should 
be.  The  thickness  of  the  bricks  influences  greatly  the  rela- 
tive amount  of  free  space  and  filled  space,  and  the  rate  at 
which  the  generator  heats  up  or  cools  off.  In  a  regenerator 
of  given  length  and  cross-section  closer  packing  of  the  bricks 
gives  more  heat  absorbing  surface,  increases  the  velocity  of  the 
gases  and  diminishes  the  cross-sectional  area  of  each  passage 
and  of  the  sum  of  all  the  passages;  some  of  these  factors  in- 
crease the  efficiency  of  the  regenerator,  others  tend  to  decrease 
it,  and  there  are,  therefore,  several  independent  variables  to  be 
considered  in  finding  the  best  arrangement  for  highest  effi- 
ciency. A  numerical  solution  is  indeed  a  possibility,  but  is 
too  involved  for  an  elementary  presentation  of  the  subject. 

Relative  Sizes. — The  relative  sizes  of  gas  and  air  regenera- 
tors is  a  question  of  importance  which  admits  of  easy  solu- 


384  METALLURGICAL  CALCULATIONS. 

tion  by  calculation.  So  far  we  have  treated  the  pair  of  re- 
generators together,  and  discussed  the  sum  of  their  cross-sec- 
tions as  deduced  from  the  volume  of  products  passing  through 
at  an  assumed  maximum  temperature  and  allowable  velocity. 
The  regenerators  at  one  end  of  a  furnace  are,  however,  usually 
divided  into  a  pair  or  set,  one  for  heating  gas  and  the  other 
for  heating  air.  This  is  not  usual  where  natural  gas  is  used 
because  of  the  deposition  of  soot  in  the  regenerator  by  the 
latter  when  it  is  heated,  but  nine  out  of  ten  open-hearth  fur- 
naces preheat  their  gas  as  well  as  the  air.  The  heating  capacity 
of  the  regenerators  should  be  divided  in  proportion  to  the 
calorific  capacities  of  the  gas  and  air  simultaneously  heated. 
The  problem  is  therefore  to  find  the  heat  capacity  per  degree 
of  the  gas  and  air  used,  or,  more  exactly,  the  total  heat  capacity 
of  each  of  these  between  the  temperature  at  which  they  enter 
the  regenerators  and  that  at  which  it  is  desired  that  they  should 
enter  the  furnace. 

Problem  69. 

An  open-hearth  furnace  uses  producer  gas  containing,  by 
volume,  at  it  reaches  the  regenerators: 

CO 26.97  per  cent. 

CO2 4.37 

CH4 0.33 

H2 13.00 

NH3 0.21 

H2S 0.10 

N2 54.01 

Air . 1.03 

Each  cubic  meter,  measured  at  20°  C.  and  720  m.  m.  baro- 
metric pressure,  is  accompanied  by  73.22  grams  of  moisture,  as 
determined  by  drawing  through  a  drying  tube  and  weighing  the 
moisture.  The  air  used  is  at  20°  C.,  720  m.m.  barometer,  and 
three-quarters  saturated  with  moisture.  A  maximum  of  10  per 
cent,  more  air  is  used  than  is  theoretically  necessary  to  com- 
pletely burn  the  gas  (assuming  NH3  to  burn  to  H2O  and  NO2, 
and  allowing  for  the  air  already  present  in  the  gas).  The 
gas  and  air  may  be  both  assumed  as  coming  to  the  regenerators 
at  20°  C.,  and  to  be  heated  in  the  regenerators  to  1200°  C. 


THE  OPEN-HEARTH  FURNACE.  385 

Required. — (1)  The  relative  volumes  of  gas  and  air  passing 
through  the  gas  and  air  regenerators. 

(2)  The  total  amounts  of  heat  necessary  to  be  furnished  to 
each,  per  cubic  meter  of  gas  used. 

(3)  The  relative  sizes  of  the  two  regenerators. 

Solution. — (1)  Taking  1  cubic  meter  of  the  dry  gas,  as  rep- 
resented by  the  analysis,  the  combustion  of  its  combustible  in- 
gredients is  represented  by  the  equations: 

2CO    +02    =  2CO2 

CH4    +  2O2  =  CO2  +  2H2O 

2H2    +O2    =2H2O 

4NH3  +  7O2  =6H2O  +  4NO2 

2H2S  +3O2  =  2H2O  +  2SO2 

And  since  molecules  represent  volumes,  each  unit  volume  of 
CO,  CH4,  H2,  NH3  and  H2S  is  seen  to  require  respectively  0.5, 
2,  0.5,  1.75,  or  1.5  volumes  of  oxygen.  The  1  cubic  meter 
of  dry  gas  therefore  requires  oxygen  as  follows: 

CO  0.2697X0.5    =  0.1349m3 
CH4  0.0437X2.0    =  0.0874  " 
H2    0.1300X0.5    =  0.0650  " 
NH30.0021X1.75  =  0.0037  " 
H2S  0.0010X1.5    =  0.0015  " 


Total  =  0.2925  ' 

Air  needed  =  0.2925-^0.208  =  1.4062m3 
Add  10  per  cent,  excess          =  1.5468  " 
Air  present  in  gas  =  0.0103   " 

Air  to  be  supplied  =  1.5365  " 

Each  cubic  meter  of  dry  gas,  at  any  given  conditions  of  tem- 
perature and  pressure,  would  require  1.5365  cubic  meters  of  dry 
air  at  the  same  conditions  of  temperature  and  pressure.  This, 
however,  is  not  exactly  the  relation  required,  for  the  reason 
that  the  gas  is  accompanied  by  considerable  moisture,  which 
in  reality  adds  to  its  volume,  while  the  air  is  also  moist,  add- 
ing to  its  volume.  Two  corrections  must  therefore  be  ap- 
plied; first,  to  calculate  the  volume  of  the  moisture  accom- 
panying 1  cubic  meter  of  (assumed)  dried  gas;  the  second,  to 
calculate  the  volume  of  moisture  accompanying  1.5365  cubic 


386  METALLURGICAL  CALCULATIONS. 

meters  of  (assumed)  dry  air.  Assuming  our  gas  and  moist 
air  both  at  20°  and  720m.m.  pressure,  the  volume  of  moisture 
accompanying  1  cubic  meter  of  (assumed)  dry  gas  is  the  vol- 
ume of  73.22  grams  of  moisture  at  these  conditions,  which  is 


73.22-hlOOO-HO.Sl  X  2y!_t  2°  X         =  0.1024  m 


The  air  being  0.75  saturated  with  moisture  at  20°,  the  tension 
of  this  moisture  will  be  17.4X0.75  =  13  m.m.  The  air  proper 
is  therefore  under  720  —  13  =  707  m.m.  tension  instead  of 
720  m.m.  and  the  volume  of  the  moist  air  containing  1.5365 
cubic  meters  of  (assumed)  dry  gas  is  therefore: 

1.5365X^5  «  1.5648  cubic  meters. 

The  relative  volumes  of  actual  (moist)  gas  and  actual  (moist) 
air  used  are  therefore  : 

1.1024:1.5684  =  1.0000:1.419  (1) 

(2)  To  calculate  the  heat  necessary  to  raise  gas  and  air  from 
20°  to  1200°  per  cubic  meter  of  gas  used,  the  best  preliminary 
is  to  calculate  the  composition  of  the  gas;  including  its  mois- 
ture. Since  1  cubic  meter  of  (assumed)  dry  gas  is  accom- 
panied by  0.1024  cubic  meter  of  water  vapor,  the  sum  being 
1.1024  cubic  meters,  we  can  calculate  the  real  percentage  com- 
position to  be: 

CO  .............................  24,47  per  cent. 

CO2  .............................  3.96 

CH4  .......................  .....  0.30 

H2  ..............................  11.7§ 

NH3  .....................  .......  0.19 

H2S  ............................  0.09 

N2  ..............................  48.99 

Air  .............................   0.93 

H20  ............................   9.29 

Of  the  above  quantities  the  (assumed)  dry  gas  in  1  cubic  meter 
of  (actual)  moist  gas  is  0.9071  c.m.  The  air  used  for  its  com- 
bustion will  therefore  be: 


THE  OPEN -HEARTH  FURNACES  387 

0.9071X1.5365  =  1.3938  tn3  dry  air. 
0.9071X1.5648  =  1.4080  m3  moist  air. 
1.4080—1.3938  =  0.0142  m3  of  moisture. 

The  heat  required  by  the  1  cubic  meter  of  (actual)  moist  gas 
is  found  as  follows: 

Volume  XMean  Specific  Heat 
20°— 1200° 


CO     0.2447 

H2      0.1179 

'  X  0.3359 

=  0.2895  Calories. 

N2      0.4899 

Air     0.0093 

CO*    0.0396     X  0.6384 

=  0.0253      " 

H2O  0.0929     X  0.5230 

=  0.0486      " 

CH4  0.0030     X  0.6484 

=  0.0019      " 

NH3  0.0019     X  0.5752 

=  0.0011       " 

H2S    0.0009     X  0.5230 

=  0.0005      " 

Mean  cal.  capacity  per  1°  =  0.3669      " 
Total  calorific  capacity  20°— 1200° 
0.3669X1180  =  432.9  Calories. 

The  calorific  capacity  of  the  moist  air  simultaneously  heated 
through  the  same  range  will  be. 

Air     1.3938X0.3359  =  0.4682  Calories. 
H2O  0.0142X0.5230  =  0.0074       " 

Sum  =  0.4756       « 

Total  calorific  capacity  20°— 1200° 
0.4756X1180  =  561.2  Calories. 

The  air  regenerator  should  therefore  have  561.2-5-432.9  = 
1.30  times  the  heating  power  or  cross-section  of  the  gas  re- 
generator; i.  e.,  30  per  cent.  more.  Or  the  combined  capacity 
of  the  pair  of  regenerators  should  be  divided  so  as  to  give  57 
per  cent,  to  the  air  regenerator  and  43  per  cent,  to  the  gas 
regenerator.  In  ordinary  practice  it  is  usual  to  allow  about  60 


388  METALLURGICAL  CALCULATIONS. 

and  40  per  cent,  respectively;  it  is  better  to  calculate  ahead  for 
the  specific  case  in  hand,  if  the  composition  of  the  gas  to  be 
used  is  known. 

VALVES  AND  PORTS. 

No  very  exact  rule  can  be  given  as  to  the  size  of  the  gas 
and  air  valves,  or  those  leading  the  products  to  the  chimney. 
If  made  too  large  they  are  cumbersome  to  operate  and  apt 
to  warp;  if  made  too  small  they  give  undue  obstruction  to  the 
flow  of  gas.  A  general  rule  is  to  calculate  the  free  opening, 
such  as  to  give  the  gases  passing  through  a  velocity  between 
3  and  5  meters  (10  and  16  feet)  per  second,  allowing,  of  course, 
for  the  average  temperature  of  the  gas  or  air  products  of  com- 
bustion passing  through  them.  It  may  be  remarked  that 
while  water-seal  valves  are  very  convenient,  the  water  is  evap- 
orated where  in  contact  with  gas  or  air,  and  diminishes  the 
heating  efficiency  of  the  furnace,  the  use  of  a  non- volatile 
oil  or  a  fine  sand  would  appear  perferable  to  water. 

The  ports  are  a  very  important  part  of  the  furnace,  and  may 
be  designed  in  many  different  styles  for  various  ways  in  which 
a  furnace  is  to  be  worked.  Their  cross-section,  however,  can 
be  calculated  when  we  know  the  volume  of  gas  or  air  leaving 
the  regenerators  and  their  temperature,  or  the  volume  of  the 
products  of  combustion  entering  the  regenerators  and  their 
temperature.  They  should  be  so  designed  that  the  velocity  of 
the  gases  through  them  is  not  over  20  meters  per  second,  while 
10  meters  per  second  is  a  better  velocity  to  use.  A  long  fur- 
nace can  admit  of  higher  velocities  at  the  ports  than  a  short 
one;  but  in  any  case  the  higher  the  velocity  the  farther  com- 
plete combustion  will  occur  from  the  ports,  and  if  the  velocity 
is  too  high  for  the  length  of  the  furnace  combustion  may  even 
be  continued  in  the  opposite  regenerators  and  less  than  the 
maximum  occur  in  the  furnace.  This  is  a  condition  to  be 
scrupulously  avoided  if  possible. 

Problem  70. 

Producer  gas  of  the  following  composition: 

CO 24.47  per  cent.         NH3 0.19  per  cent. 

CO2 3.96        "  N2 48.99 

CH4 0.30        "  Air 0.93 

H2 11.79        "  H20 9.29 

H2S..  .   0.09 


THE  OPEN-HEARTH  FURNACE.  389 

is  burned  with  1.408  times  its  volume  of  moist  air  (see  Prob- 
lem 69).  The  furnace  treats  50  metric  tons  of  steel  in  12  hours, 
using  17.5  tons  of  coal  in  the  producers,  from  which  15  tons 
of  carbon  pass  into  the  gas.  The  gas  and  air  pass  out  of  the 
regenerators  at  1200°,  and  the  products  of  combustion  (as- 
sumed complete)  pass  into  the  opposite  regenerators  at  1400°. 
Assume  a  maximum  velocity  of  the  hot  gas  and  air  as  10  meters 
per  second,  as  they  pass  through  the  ports. 

Required. — (1)  The  volume  of  gas  and  air  at   20°  C.   and 
720  m.m.  barometer  used  by  the  furnace  per  second. 

(2)  The  areas  of  the  gas  and  air  ports. 

(3)  The  velocity  of  the  products  entering  the  opposite  ports. 
Solution. — (1)   The  carbon  in  1  cubic  meter  of  the  gas  at 

standard  conditions  is 

CO  0.2447 
CO2  0.0396 
CH4  0.0030 


0.2873X0.54  =  0.1551kg. 
Gas  used  in  12  hours  (standard  conditions)  : 

15,000^-0.1551  =  96,710m3 
Per  second  =  2.24  m3 
Gas  used  at  20°  C.  and  720  m.m.  : 


2.24  X  X  y     =  2-53  m3  Per  second-  (D 

• 

Air  used  (standard  conditions): 

2.24X1,408  =  3.15m3 
Air  used  at  20°  C.  and  720  m.m.  : 

2.53  X  1,408  =  3.56  m3  per  second.  (1) 

(2)  The  volume  of  gas  used  per  minute,  as  it  issues  from  the 
ports  at  1200°,  is 

' 


and  of  air       3.15X         "  X     "     =  17.9  " 

and  assuming  a  maximum  velocity  for  each  of  10  meters  per 
second,  the  areas  of  the  ports  must  be: 


390  METALLURGICAL  CALCULATIONS. 

Gas  ports 1 . 28  m2 

Air  ports 1.79    " 


Sum 3.07  (2) 

(3)  There  is  usually  contraction  when  gases  burn,  the  pro- 
ducts having  less  volume  than  the  gas  and  air  used.  In- 
specting the  equations  of  combustion  of  CO,  CH4,  H2,  H2S  and 
NH3  given  in  Problem  69,  we  can  construct  the  following  table 
of  relative  volumes  concerned  and  the  ensuing  contraction: 

CO   CH4  H2  IPS  NH3 

Volumeused 1.01.0  1.0  1.0  1.0 

Oxygen  used 0.52.0  0.5  1.5  1.75 

Gases  combining 1.53.0  1.5  2.5  2. 75 

Volume  of  products ...1.03.0  1.0  2.0  2.5 

Contraction 0.5  0.0  0.5  0.5  0.25 

Using  1  cubic  meter  of  producer  gas  the  contraction  resulting 
from  its  combustion  with  an  excess  of  air  is 

CO  0.2447X0.5   =  0.12235m3 

CH4  =  0.00000    " 

H2  0.1179X0.5   =  0.05895    " 

H2S  0.0009X0.5   =  0.00045    " 

NH3  0.0019X0.25=  0.00050   " 


Total  contraction  =  0.1822 

Since  the  volume  of  gas,  plus  air  used,  is  2.408  m3,  the  volume 
of  the  products,  at  standard  conditions,  is 

2.408-0.182  =  2.226  m3 

per  cubic  meter  of  gas  used  under  standard  conditions.  The 
volume  of  products  per  minute,  at  standard  conditions,  is, 
therefore, 

2.226X2.24  =  4.986  m3. 

And  at  1400°  and  720  m.m.  pressure: 


THE  OPEN-HEARTH  FURNACE.  391 

Since  the  sum  of  the  area  of  gas  and  air  ports  is  2.91  m2,  the 
velocity  of  the  products  in  these  ports  will  be 

32.7-:-3.07  =  10.7  m.  per  second.  (3) 

In  both  the  calculations  of  the  size  of  the  ports  and  the 
velocity  of  the  products  we  have  assumed  the  tension  of  the 
gas,  air  or  products  in  the  ports  to  be  the  prevailing  atmos- 
pheric tension.  This  may  or  may  not  be  exactly  true,  because 
the  air  or  gas  may  be  under  a  slightly  less  tension,  being  drawn 
into  the  furnace  by  the  stack  draft.  If  the  pressure  inside 
the  furnace,  with  doors  closed,  is  greater  or  less  than  atmos- 
pheric pressure,  the  tension  of  the  gases  in  the  entrance  ports 
will  be  correspondingly  greater  or  less  than  the  atmospheric 
pressure,  while  the  tension  of  the  products  will  probably  always 
be  less  than  atmospheric  pressure,  because  of  the  stack  draft. 
Under  ordinary  conditions  these  corrections  are  too  small  to 
need  to  be  taken  into  consideration. 

LABORATORY  OF  FURNACE. 

The  laboratory  consists  of  the  open  space  enclosed  between 
the  hearth,  sides,  ends  and  roof.  Its  dimensions  vary  with  the 
intended  capacity  of  the  furnace  and  the  ideas  of  the  designer. 
If  a  hearth  is  to  contain,  say,  50  tons  of  melted  steel,  which 
weighs  some  7  tons  per  cubic  meter  (425  pounds  per  cubic 
foot),  there  will  be  contained  in  the  furnace  at  one  time  7  cubic 
meters,  or  260  cubic  feet  of  steel.  The  deeper  this  lies  the 
more  slowly  it  will  be  heated  or  oxidized  by  the  flame,  and  there- 
fore there  is  a  limiting  depth  of,  say,  50  centimeters,  or  20 
inches,  which  it  is  not  advisable  to  exceed,  while  a  more  shallow 
bath  will  result  in  faster  working.  Assuming  a  depth  of  40 
centimeters  (16  inches),  the  volume  divided  by  the  depth  will 
give  the  area  of  the  bath: 


7^-0.4    =  17.5  square  meters 
260-^-1.25  =  208  square  feet. 


We  can  then  either  choose  a  convenient  width,  consistent 
with  a  practicable  roof  span,  and  derive  the  length,  or  choose 
a  length  and  derive  the  width,  or  choose  a  certain  ratio  of 
length  to  width,  and  derive  both.  If  the  width  is  3  meters  the 


392  METALLURGICAL  CALCULATIONS. 

length  must  be  5.8;  if  the  length  is  assumed  5  meters,  the  width 
is  3.5;  if  the  ratio  of  length  to  width  is  2  to  1,  the  length  figures 
out  5.92  meters  and  the  width  2.96.  These  dimensions  are 
those  of  the  bath  of  metal,  and  each  should  be  increased  by 
at  least  1  meter  to  get  the  area  of  the  hearth  inside  the  walls, 
thus  allowing  0.5  meter  clear  space  all  around  the  metal. 

If  a  furnace  is  short  it  should  be  wide  and  the  roof  high, 
in  order  to  give  cross-sectional  area  and  thus  diminish  the 
velocity  of  the  gases  over  the  hearth.  The  gases  attain  their 
maximum  temperature  in  the  laboratory,  theoretically,  some 
1700°  to  1900°,  and  their  velocity  depends  solely  on  the  vertical 
cross-sectional  area  of  the  laboratory  or  body  of  the  furnace. 
In  Problem  70,  for  instance,  about  5  cubic  meters  of  products 
of  combustion  (measured  at  standard  conditions)  pass  through 
the  furnace  per  second.  At  1800°  this  volume  would  be  38 
cubic  meters,  and  if  the  laboratory  were  4.5  meters  wide  by 
1.5  meters  high  above  the  level  of  bath,  there  would  be  7.25 
square  meters  of  cross-section,  and  the  velocity  of  the  gases 
would  be  38 -J-  7.25  =  5.2  meters  per  second.  This  would 
allow  barely  1  second  for  the  hot  gases  to  pass  over  the  bath, 
which  would  result  in  a  low  rate  of  heating  and  probable  in- 
complete combustion,  for  the  gas  can  only  burn  as  it  gets  mixed 
with  air,  and  it  is  hardly  likely  that  100  per  cent,  of  it  would 
get  mixed  with  air  and  consumed  in  1  second.  Such  could 
only  be  attained  by  sub-division  of  the  gas  and  air  and  very 
intimate  mixture  at  the  ports.  Much  of  the  economy  undoubt- 
edly attained  by  raising  the  roof  of  open-hearth  furnaces  is  due 
to  the  slowing-up  of  the  gas  currents  in  the  laboratory,  though 
it  is  usually  ascribed  to  avoidance  of  contact  of  flame  and 
bath,  increased  heating  by  radiation,  etc.  In  the  writer's  opinion 
the  raising  of  the  roof  from  1  to  2  meters,  let  us  say,  thus  doub- 
ling the  vertical  cross-sectional  area,  cutting  in  half  the  velocity 
of  the  gases  through  the  furnace,  and  doubling  the  period  in 
which  they  are  able  to  combine,  and  to  radiate  or  impart  heat 
to  the  furnace  walls  and  charge — is  the  principal  reason  for 
the  increased  economy  observed. 

An  equally  important  improvement  is  lengthening  the  dis- 
tance between  ports.  There  is  a  limit  to  the  width  of  the 
furnace,  set  by  the  practicable  arch  for  the  roof;  there  is  also 
a  limit  to  the  height  of  roof,  set  by  the  increasing  distance  of 


THE  OPEN-HEARTH  FURNACE.  393 

the  gases  from  the  hearth ;  when  both  these  factors  have  reached 
their  maximum,  further  efficiency  of  utilization  of  the  heat 
of  combustion  can  only  be  secured,  as  far  as  the  body  of  the 
furnace  is  concerned,  by  lengthening  the  hearth.  There  is 
no  mechanical  limit,  and  in  every  case  the  distance  between 
the  ports  and  the  velocity  of  the  gases  should  be  such  that 
complete  combustion  takes  place  in  the  furnace  laboratory 
before  the  products  pass  into  the  regenerators. 

Problem  71. 

H.  H.  Campbell  gives  in  the  Transactions  of  the  American 
Institute  of  Mining  Engineers,  1890,  analyses  made  at  the 
Pennsylvania  Steel  Co.'s  works,  as  follows: 

Gas  Burned,  Products, 
Entering  Furnace.        Leaving  Furnace. 

CO2 5.5  per  cent.  3.1  per  cent. 

O2 2.3        "  0.7        " 

CO 8.2        "  7.1 

CH4 7.3       "  0.0       " 

H2.... 39.8        "  11.6        " 

N2 36.9       "  77.5       " 

Required. — (1)  The  proportion  of  the  calorific  power  of  the 
fuel  developed  while  passing  through  the  body  of  the  furnace. 

(2)  The  proportion  of  the  air  necessary  for  complete  com- 
bustion which  was  used. 

Solution. — (1)  One  cubic  meter  of  the  gas  contains  the  fol- 
lowing weight  of  carbon: 

(0.055 +  0.082  + 0.073)  X  0.54  =  0.1134  kg. 
One  cubic  meter  of  products  contains: 

(0.071  + 0.031)  X  0.54  =  0.0551  kg. 
Therefore,  volume  of  products  per  1  cubic  meter  of  gas: 

0.1134 +  0.0551  =  2.06m3. 
Calorific  power  of  1  cubic  meter  of  gas: 

CO  0.082X3062  =    251  Calories 

CH4          0.073X8623  =    629 
H2  0.398X2613  =  1040 

1920 


394  METALLURGICAL  CALCULATIONS. 

Calorific  power  of  2.06  cubic  meters  of  products: 

CO  0.071X3062=    217  Calories. 

H2  0.116X2613  =    303 


520 
520X2.06  =  1071 

Heat  developed  in  the  furnace: 

1920  -  1071  =  849  Calories. 
Proportion  of  the  possible  heat  development: 

849^-1920  =  0.442  =  44.2  per  cent.  (1) 

(2)  The  1  cubic  meter  of  gas  needed,  to  burn  its  combustible 
constituents,  the  following     amount  of  oxygen: 

CO  0.082X0.5      =  0.041m3 

CH4  0.073X2.0      =  0.146   " 

H2  0.98  X0.5       =  0.199  " 

Sum   =  0.386   " 
Oxygen  present  in  gas        =  0.023   ' 

Oxygen  needed  from  air     =  0.363   " 
Air  =  0.363^0.208  =  0.745  " 

The  2.06  m3  of  products  of  incomplete  combustion  require  for 
their  combustion: 

CO    0.071X2.06X0.5  =  0.0731  m3  oxygen. 
H2     0.116X2.06X0.5  =  0.1195   " 


Sum  =  0.1926   ' 
Oxygen  present  in  the  products      =  0.0070   " 

Oxygen  needed  from  the  air  =  0.1856   " 

Air  needed  to  complete  combus- 
tion =  0.892  ' 

Total  air  needed  for  complete 

combustion  =  1.745 


Air  supplied  in  the  furnace  =  0.853   ' 

Percentage  supplied: 

0.853-^1.745  -  0.489  =  48.9  per  cent.  (2) 


THE  OPEN-HEARTH  FURNACE.  395 

It  is  almost  needless  to  remark  that  with  less  than  half  the 
air  necessary  for  complete  combustion  supplied,  a  high  calorific 
intensity  of  flame  and  a  high  utilization  of  the  calorific  power 
of  the  fuel  are  impossible.  More  air  should  have  been  used 
and  the  furnace  made  longer,  so  as  to  secure  perfect  combus- 
tion in  the  furnace,  and  not  have  over  half  the  possible  de- 
velopment left  to  take  place  in  the  regenerators  or  stack. 

CHIMNEY  FLUES  AND  CHIMNEY. 

The  gases  pass  into  the  chimney  flues  at  from  150°  to  450°. 
If  their  volume  at  an  assumed  average  temperature  of,  say, 
300°  is  calculated,  they  can  be  given  an  assumed  velocity  of 
2  to  3  meters  (5  to  10  feet)  per  second,  and  thus  a  suitable  cross- 
sectional  area  of  the  chimney  flues  obtained.  The  stack  will 
work  best  with  a  velocity  of  5  meters  per  second,  and  thus  its 
cross-sectional  area  may  be  calculated.  A  height  of  25  to  30 
meters  (75  to  100  feet)  is  sufficient  for  most  furnaces. 

MISCELLANEOUS. 

Some  other  data  useful  in  figuring  up  the  dimensions  and 
running  conditions  of  modern  open-hearth  steel  furnaces  are 
the  following,  taken  mostly  from  an  article  by  H.  D.  Hess,  in 
the  Proceedings  Engineering  Club  of  Philadelphia,  January, 
1904: 

Average  coal  consumption,  in  pounds,  per  hour  per  ton  of 
metal  capacity  of  the  furnace,  55  to  80. 

Cubical  feet  of  space  in  one  pair  of  regenerators,  per  ton 
of  metal  capacity  of  the  furnace,  30  to  75. 

Cubic  feet  of  space  in  one  pair  of  regenerators  per  pound 
of  coal  consumed  per  hour,  0.5  to  1.0. 

A  correllation  and  combination  of  data  of  this  sort,  with 
details  as  to  the  actual  working  of  the  furnaces,  would  point 
the  way  towards  a  general  solution,  which  would  furnish  the 
best  condition  for  every  possible  case,  with  strict  consideration 
for  all  the  variables  involved. 


CHAPTER  XL 
THERMAL  EFFICIENCY  OF  OPEN-HEARTH  FURNACES. 

The  ordinary  open-hearth  steel  furnace  receives  cold  pig 
iron,  cold  scrap,  warm  ferro-manganese,  cold  limestone  and 
cold  ore,  it  receives  cold  air  and  moderately  warm  producer 
gas,  and  it  furnishes  melted  steel  and  slag  at  the  tapping  heat. 
The  larger  part  of  the  usefully  applied  heat  is  that  contained 
in  the  melted  steel,  for  it  must  be  melted  in  order  to  be  cast, 
and  when  once  taken  away  from  the  furnace  the  latter  is  done 
with  it. 

The  total  heat  available  for  the  purposes  of  the  furnace  and 
which  should  be  charged  against  it  consists  of  the  following 
items: 

(1)  Heat  in  warm  or  hot  charges. 

(2)  Heat  in  warm  or  hot  gas  as  it  reaches  the  furnace. 

(3)  Heat  in  warm  or  hot  air  as  it  reaches  the  furnace. 

(4)  Heat  which  could  be  generated  by  complete  combustion 
of  the  gas. 

(5)  Heat  of  oxidation  of  those  constituents  of  the  charge 
which  are  oxidized  in  the  furnace. 

(6)  Heat  of  formation  of  the  slag. 

The  items  of  distribution  of  this  total  will  be  as  follows: 

(1)  Heat  in  the  melted  steel  at  tapping. 

(2)  Heat  absorbed  in  reducing  iron  from  iron  ore. 

(3)  Heat  absorbed  in  decomposing  limestone  added  for  flux. 

(4)  Heat    absorbed   in    evaporating    any   moisture   in    the 
charges. 

These  first  three  items  constitute  the  usefully  applied  heat, 
and  their  sum  measures  the  net  thermal  efficiency  of  the  furnace. 

(5)  Heat  absorbed  in  reducing  ferric  oxide  to  ferrous  oxide. 

(6)  Heat  in  the  slag. 

(7)  Heat  lost  by  imperfect  combustion. 

396 


EFFICIENCY  OF  OPEN -HEARTH  FURNACES.  397 

(8)  Heat  in  the  chimney  gases  as  they  leave  the  furnace. 

(9)  Heat  absorbed  by  cooling  water. 

(10)  Heat  lost  by  conduction  to  the  ground. 

(11)  Heat  lost  by  conduction  to  the  air. 

(12)  Heat  lost  by  radiation. 

(1)  HEAT  IN  WARM  CHARGES. 

If  the  pig  iron  is  charged  melted  instead  of  cold  an  immense 
amount  of  thermal  work  is  spared  the  furnace,  and  it  should 
be  charged  with  all  the  heat  (reckoning  from  0°  C.  as  a  base 
line)  which  is  in  the  melted  pig  iron  as  it  runs  into  the  fur- 
nace. This  will  average  275  Calories  per  unit  of  pig  iron, 
but  should  be  actually  determined  calorimetrically  in  each 
specific  instance  wherever  possible.  The  net  thermal  efficiency 
of  the  furnace  will  figure  out  higher  with  cold  charges  than 
with  melted  pig  iron,  because,  with  a  possible  flame  tempera- 
ture of  1,900°  C.  in  the  furnace,  heat  is  absorbed  much  more 
rapidly  by  cold  charges  than  by  hot  ones,  and  a  larger  per- 
centage of  the  available  heat  will  be  thus  usefully  applied. 

Scrap  is  almost  always  charged  cold,  but  if  any  of  it  is  hot 
its  weight  and  temperature  should  be  known  and  the  amount 
of  heat  thus  brought  in  charged  against  the  furnace.  Or  a 
small  piece  may  be  dropped  into  a  calorimeter  and  its  heat 
content  per  unit  of  weight  measured  directly,  and  thus  the  heat 
in  all  the  hot  scrap  used  may  be  estimated. 

Ferro-manganese  is  often  added  cold,  but  usually  is  pre- 
heated to  cherry  redness  (about  900°)  in  another  small  fur- 
nace, in  order  that  it  may  dissolve  more  quickly  in  the  bath. 
Knowing  its  weight,  temperature  and  specific  heat,  the  heat 
which  it  brings  into  the  furnace  can  be  calculated ;  a  better  plan 
is  to  drop  a  piece  into  a  calorimeter  and  measure  the  actual 
heat  in  a  sample  of  it. 

Limestone  and  ore  are  almost  invariably  put  into  the  fur- 
nace cold.  If  used  warm  the  heat  in  them  can  be  determined 
by  the  methods  just  described. 

(2)  HEAT  IN  THE  GAS  USED. 

By  this  is  meant,  not  the  heat  in  the  gas  after  it  is  heated 
by  the  regenerators,  but  its  sensible  heat  as  it  reaches  the 
furnace.  This  applies  only  to  furnaces  where  the  producers 


398  METALLURGICAL  CALCULATIONS. 

or  gas  supply  are  independent  of  the  furnace.  Where  the  pro- 
ducers are  an  integral  part  of  the  furnace  it  is  impracticable 
to  consider  them  separately  from  the  furnace,  and  the  efficiency 
of  the  whole  plant,  including  the  producers,  must  be  consid- 
ered together.  But  where  the  gas  supply  from  whatever  source 
comes  to  the  furnace  from  outside,  and  reaches  the  furnace 
warm,  its  sensible  heat  is  to  be  charged  against  the  furnace  as 
part  of  the  heat  which  the  furnace  must  account  for.  If  the 
gas  comes  from  producers  its  amount  is  satisfactorily  found 
from  the  known  weight  of  carbon  gasified  per  hour,  or  per 
furnace  charge,  and  the  weight  of  carbon  contained  in  unit 
volume  of  gas,  as  calculated  from  its  analysis.  If  gas  comes 
from  a  common  main  which  supplies  several  furnaces,  or  is 
simply  natural  gas,  its  amount  can  only  be  roughly  estimated 
by  measuring  the  area  of  the  gas  supply  pipe  or  flue  and  meas- 
uring the  velocity  of  flow  by  a  pressure  gauge  or  Pitot  tube 
or  anemometer.  None  of  these  methods  just  mentioned  are 
satisfactorily  accurate,  and  there  is  great  need  of  simple  methods 
for  determining  accurately  the  flow  of  gases  in  flues  or  pipes,. 
If  the  velocity  of  warm  gas  is  determined  suitable  correction 
for  its  temperature  must  be  made  to  reduce  it  to  volume  at 
standard  conditions. 


(3)  HEAT  IN  THE  AIR  USED. 

If  the  air  coming  to  the  furnace  is  warm  its  sensible  heat 
must  be  charged  against  the  furnace.  If  the  air  is  warmed, 
however,  before  it  goes  into  the  regenerators  by  waste  heat 
from  the  furnace  itself,  then  its  sensible  heat  should  not  be 
charged  against  the  furnace,  because  that  would  amount  to 
charging  the  furnace  twice  with  this  quantity  of  heat.  Such 
preheating  it  in  reality  only  a  part  of  the  regenerative  princi- 
ple, even  though  it  may  not  be  done  in  regenerators,  but,  for 
instance,  by  circulating  air  around  a  slag-pot  or  through  the 
hollow  walls  of  the  furnace.  If  the  air  used  is  moist  its  mois- 
ture should  not  be  omitted  in  the  calculation. 

The  amount  of  air  used  is  best  determined  by  a  comparison 
of  the  analyses  of  gas  and  chimney  products,  and  a  calculation 
based  on  the  carbon  contents  of  each  and  the  known  volume 
of  gas  used. 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  399 

(4)   HEAT  OF  COMBUSTION. 

Under  this  head  comes  the  principal  item  of  heat  available 
for  the  furnace.  In  reckoning  it  we  should  calculate  the  total 
heat  which  could  be  generated  by  the  perfect  combustion  of 
the  gas  used,  to  CO2,  N2  and  H2O  vapor.  If  there  is  in  reality 
imperfect  combustion,  as  is  shown  by  analysis  of  the  chimney 
gases,  that  is  a  defect  of  operation  of  the  furnace  which  should 
be  written  down  against  it.  Problem  71  showed  an  actual  case 
in  which  there  was  very  incomplete  combustion  in  the  body  of 
the  furnace,  but  where  combustion  was  afterwards  completed 
in  the  regenerators.  In  such  a  case  the  same  principle  applies; 
the  furnace  must  be  charged  with  the  total  calorific  power  of 
the  fuel  used,  and  incomplete  combustion  can  be  charged 
against  the  furnace  as  a  whole  only  on  the  basis  of  uncon- 
sumed  ingredients  in  the  chimney  gases — the  products  finally 
rejected  by  the  furnace.  If  there  is  poor  combustion  in  the  body 
of  the  furnace  and  combustion  is  only  completed  in  the  re- 
generators, the  furnace  will  not  give  as  high  net  thermal  effi- 
ciency as  if  combustion  were  complete  above  the  hearth. 

(5)  OXIDATION  OF  THE  BATH. 

The  oxidation  of  carbon,  iron,  silicon,  manganese  and  some- 
times phosphorus  and  sulphur,  add  a  not  inconsiderable  amount 
to  the  heat  resources  of  the  furnace.  Carbon  should  be  burnt 
to  CO2,  iron  is  usually  oxidized  to  FeO,  manganese  to  MnO, 
silicon  to  SiO2,  phosphorus  to  P2O5,  and  sulphur  to  SO2.  All 
of  these  oxidations  generate  heat,  and,  moreover,  heat  which 
should  be  very  efficiently  utilized,  being  produced  in  direct 
contact  with  the  metallic  bath ;  it  should  all  be  charged  against 
the  furnace  as  part  of  its  available  heat. 

(6)  FORMATION  OF  SLAG. 

The  metallic  oxides,  produced  unite  with  each  other,  and 
with  the  lime  and  silica  of  ore  used  and  lining  of  the  hearth  to 
produce  the  slag,  the  heat  of  formation  of  which  can  be  calcu- 
lated and  counted  in  as  available  heat. 

(1)  HEAT  IN  MELTED  STEEL. 

This  is  a  large  item  in  the  work  done  by  the  furnace ;  in  fact, 
usually  the  largest  single  item.  It  should  be  determined 


400  METALLURGICAL  CALCULATIONS. 

calorimetrically  when  possible;  if  this  is  not  done  its  tem- 
perature should  be  known  and  its  composition,  in  order  to 
compare  it  with  the  calorimetric  experiments  of  others,  and 
thus  derive  a  probable  value  for  its  heat  contents.  Not  many 
experimental  values  in  this  line  have  so  far  been  published, 
and  a  very  much  needed  investigation  is  one  upon  the  total 
heat  in  melted  steels  of  different  compositions  at  different 
temperatures.  Values  from  275  to  350  Calories  per  kilogram 
have  been  observed. 

(2)  HEAT  OF  REDUCTION  OF  IRON  FROM  ORE. 
This  is  an  item  which  appears  whenever  ore  is  used  to  facili- 
tate oxidation  of  the  bath.  The  weight  and  composition 
of  the  charges  and  the  products  will  easily  show  how  much 
iron  has  been  reduced.  The  ore  used  is  almost  always  hema- 
tite, less  frequently  magnetite;  hydrated  iron  oxides  are  not 
used  for  obvious  reasons.  The  heat  absorbed  is  1,746  Calories 
per  kilogram  of  iron  reduced  from  Fe2O3  and  1,612  Calories 
per  kilogram  from  Fe3O4. 

(3)  DECOMPOSITION  OF  LIMESTONE  FLUX. 
If  limestone  is  charged  raw,  as  is  usually  done  in  order  to 
avoid  the  dusting  caused  by  using  burnt  lime,  then  the  furnace 
is  called  upon  to  burn  this  limestone  in  place  of  the  lime  kiln. 
The  heat  absorbed  may  be  taken  as  either 

451  Calories  per  kilogram  of  CaCO3  decomposed. 
1,026  Calories  per  kilogram  of  CO2  driven  off. 
806  Calories  per  kilogram  of  CaO  produced. 

(4)  EVAPORATION  OF  MOISTURE  IN  THE  CHARGES. 
If  the  ore,  flux,  scrap  or  ore  are  put  into  the  furnace  wet 
their  moisture  must  be  evaporated.  The  correct  figure  for  this 
evaporation  is  the  latent  heat  at  ordinary  temperatures,  viz.: 
606.5  Calories  per  kilogram.  This  allows  for  the  heat  re- 
quired to  convert  into  cold  vapor  of  water,  and  puts  the  H2O 
thereafter  upon  the  same  basis  as  all  the  other  gas  going  out 
of  the  furnace.  The  chimney  gases  carry  out  sensible  heat, 
and  the  H2O  in  them  can  be  calculated  as  carrying  out  a  cer- 
tain amount  of  heat  as  gas,  reckoning  from  zero,  and  thus 
the  correct  chimney  loss  obtained.  It  is  incorrect  either  to 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  401 

charge  the  latent  heat  of  vaporization  as  chimney  loss  or  to 
charge  the  sensible  heat  of  the  water  vapor  in  the  chimney 
gases  to  heat  absorbed  in  evaporating  water  in  the  furnace. 
It  is  also  incorrect  to  do  as  is  frequently  done,  viz.:  to  calcu- 
late the  heat  required  to  evaporate  the  moisture  to  water 
vapor  at  100° — 637  Calories — and  say  that  this  is  the  heat  to 
evaporate  the  moisture.  With  almost  no  moisture  in  the  gases, 
the  moisture  of  the  charges  would  commence  to  evaporate  at 
once,  while  they  were  yet  cold,  and  the  moisture  is  no  more 
evaporated  at  100°  or  to  vapor  at  100°  than  it  is  to  200°  or 
500°.  The  only  safe  course  is  to  confine  the  evaporation  heat 
to  that  necessary  to  convert  the  moisture  into  cold  vapor,  and 
let  its  sensible  heat  as  it  escapes  as  vapor  at  any  other  tem- 
perature be  reckoned  in  with  the  sensible  heat  of  the  chimney 
gases. 

(5)  REDUCTION  OF  ORE  INTO  THE  SLAG. 
While  considerable  of  the  iron  in  the  ore  used  is  reduced  to 
the  metallic  state,  yet  often  the  larger  part  is  reduced  merely 
to  the  state  of  FeO,  and  as  such  goes  into  the  slag.  The  amount 
so  reduced  can  be  determined  by  subtracting  the  iron  reduced 
from  ore  from  the  total  iron  in  the  ore  used;  the  differences 
gives  the  iron  from  the  ore  going  into  the  slag  as  FeO.  The 
weight  of  FeO  corresponding  is  then  easily  calculated.  The 
heat  absorbed  in  this  partial  reduction  is: 

446  Calories  per  kilogram  of  FeO  reduced  from  Fe208. 

341  Calories  per  kilogram  of  FeO  reduced  from  Fe3O4. 

(6)  HEAT  IN  SLAG. 

This  is  usually  a  small  item  in  open-hearth  practice,  but  may 
amount  to  a  very  considerable  one  in  the  method  of  running 
with  large  ore  charges,  as  in  the  Monell  process.  The  varia- 
tions of  composition  of  the  slag,  and  especially  in  the  tempera- 
ture at  which  it  is  run  off,  are  so  large  that  the  heat  in  the 
slag  should  always  be  determined  calorimetrically  for  each 
specific  case.  If  assumptions  have  to  be  made,  450  to  550 
Calories  per  kilogram  of  slag  would  be  assumed.  The  weight 
of  slag  is  seldom  taken,  although  it  could  in  most  cases  be 
done  if  desired.  If  the  weight  is  not  known  it  may  be  calcu- 
lated from  the  known  weight  of  either  iron,  manganese  or 


402  METALLURGICAL  CALCULATIONS. 

phosphorous  going  into  it,  as  seen  from  the  balance  sheet  and 
the  percentages  of  these  elements  in  the  slag  as  shown  by  ana- 
lysis. 

(7)  Loss  BY  IMPERFECT  COMBUSTION. 

This  is  based  upon  the  unconsumed  ingredients  of  the  chim- 
ney gases,  as  shown  in  an  analysis.  From  this  the  calorific 
power  of  the  unburnt  gases  in  1  cubic  meter  can  be  calculated. 
If  then  we  know  the  volume  of  chimney  gases  per  unit  of  charge, 
we  get  the  heat  loss  by  imperfect  combustion  per  unit  of  charge. 
The  volume  of  chimney  gas  is  found  by  means  of  the  carbon 
in  it,  which  must  all  come  from  the  carbon  in  the  gas  used, 
plus  the  carbon  oxidized  out  of  the  charges,  plus  the  carbon 
of  CO2,  driven  off  raw  limestone  used  as  flux.  Or,  putting 
it  in  another  way,  the  total  carbon  going  into  the  furnace 
in  any  form,  less  the  carbon  in  finished  steel,  must  give  the 
carbon  in  the  chimney  gases.  This  divided  by  the  weight  of 
carbon  in  unit  volume  of  chimney  gas  gives  'the  volume  of 
the  latter,  per  whatever  unit  of  charge  is  used  as  the  basis 
of  calculations.  This  volume  times  the  calorific  power  of  unit 
volume  of  chimney  gas,  gives  the  total  heat  lost  by  imperfect 
combustion. 

(8)  SENSIBLE  HEAT  OF  CHIMNEY  GASES. 

The  temperature  of  these  gases  should  be  taken  as  they 
enter  the  chimney  flue.  Their  amount  is  determined  as  ex- 
plained under  the  previous  heading.  The  water  vapor  con- 
tained should  not  be  overlooked,  being  reckoned  simply  as 
vapor  or  gas  in  exactly  the  same  category  as  the  other  gases. 
The  analysis  of  the  chimney  gases  being  usually  given  on  dried 
gas,  a  separate  determination  of  the  moisture  carried  per  unit 
volume  of  such  dried  gas  is  necessary.  If  this  is  not  done  an 
approximation  can  be  made  by  considering  all  the  hydrogen 
in  the  gas  burned  to  form  water  vapor,  and  add  in  the  moisture 
of  the  air  used  and  the  moisture  in  the  charge. 

(9)  HEAT  LOST  IN  COOLING  WATER. 

This  is  a  very  variable  amount,  and  must  be  determined  for 
each  furnace  by  measuring  the  amount  of  water  used  per  unit 
of  time  and  its  temperature  before  reaching  and  after  leaving 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  403 

the  furnace.  Doors  are  frequently  water-cooled,  also  ports, 
where  the  heat  is  fiercest,  and  sometimes  a  ring  around  the 
hearth  at  the  slag  line. 

(10)  Loss  BY  CONDUCTION  TO  THE  GROUND. 

This  is  a  quantity  extremely  difficult  to  measure  or  to  esti- 
mate. If  a  closed  vessel  filled  with  water  were  put  into  the 
foundations  the  rate  at  which  its  temperature  rose  might  give 
some  idea  of  the  rate  at  which  heat  passed  in  that  direction  per 
unit  of  surface  contact.  At  present,  lacking  all  reliable  data, 
we  must  put  this  item  in  the  "  by  difference  "class. 

(11)  Loss  BY  CONDUCTION  TO  THE  AIR. 

This  is  an  amount  which  can  be  observed  and  calculated  with 
some  approach  to  satisfaction.  The  sina  qua  non  for  this  par- 
pose  is  a  FeYy  radiation  pyrometer,  by  which  the  temperature 
of  the  outside  of  the  furnace  at  different  parts  can  be  accurately 
determined.  Then  the  velocity  of  the  air  blowing  against  the 
furnace,  if  it  is  in  a  current  of  air,  is  observed  with  a  wind 
gauge,  and  its  temperature  before  reaching  the  furnace.  With 
these  data  and  by  the  methods  of  calculation  before  explained 
in  these  calculations  (Part  I,  Chap.  VIII)  the  heat  lost  to  the 
air  may  be  calculated. 

(12)  RADIATION  Loss. 

Having  determined  the  temperature  of  the  outer  surface  of 
the  furnace  and  measured  its  extent,  as  above  explained,  the 
radiation  loss  can  also  be  calculated,  knowing  the  mean  tem- 
perature of  the  surroundings,  by  the  principles  of  radiation, 
having  due  regard  to  the  nature  of  the  radiating  surface.  Tables 
of  specific  radiation  capacity  of  different  substances  (fire-brick, 
stone,  iron)  will  be  found  at  the  reference  just  given  above. 

Problem  72. 

Jiiptner  and  Toldt  (Generatoren  und  Martinofen,  p.  73)  ob- 
served the  following  data  with  regard  to  an  open-hearth  steel 
furnace  charge: 

Weight  of  cold  charges,  at  26°  C 3,745  kg. 

Weight  of  hot  charges,  at  700°  C 1,700   " 

Total  weight  of  charge 5,445   " 


404  METALLURGICAL  CALCULATIONS. 

Average  composition  of  charge                      C  =  1.07  per  cent. 

Si  =  0.50       " 

Mn  =  1.33 

Coal  gasified  in  producers 1,980  kg. 

Carbon  gasified  from  coal 47.13  per  cent. 

Average  composition  of  dried  gas  used CO2  3.81        " 

O2  0.98 

CO  23.82 

CH4  0.42       " 

H2  8.75       " 

N2  62.22       " 

Moisture  accompanying  each  m3  of  gas =  82  grams. 

Temperature  of  gas  reaching  furnace. 165°  C. 

Temperature  of  air  used 26°  C 

Moisture  accompanying  each  m3  of  air 12  grams. 

Barometer 717  m.m. 

Steel  produced 5,191  kg. 

Cpmposition  of  steel                                       C  =  0.12  per  cent. 

Si  =  0.04 

Mn  =  0.19       " 

Temperature  of  steel  when  tapped 1410°  C. 

Heat  in  1  kg.  steel,  by  calorimeter  (to  0°  C.) .  .  277  Cal. 

Composition  of  slag                                        SiO2  45.65  per  cent. 

FeO  33.60       " 

MnO  18.21 

CaO  2.54       " 

Weight  of  slag. . . 425  kg. 

Temperature  of  slag  when  issuing 1,410°  C. 

Heat  in  1  kg.  slag,  by  calorimeter  (to  0°) 560  Cal. 

Composition  of  the  stack  gases  (dry)            CO2  11.12  per  cent. 

O2  6.78       " 

N2  82.10       " 

Temperature  of  the  gases  in  chimney  flue 500°  C. 

Requirements: 

(1)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace. 

(2)  A  balance  sheet  of  the  heat  available  and  its  distribution. 

(3)  What  excess  of  air  above  what  is  necessary  for  complete 
combustion  is  used,  and  what  per  cent,  of  all  the  available 
heat  of  the  furnace  is  thereby  lost? 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES. 


405 


(4)  What  is  the  thermal  efficiency  of  the  furnace? 
Solution:     (1) 

BALANCE  SHEET  OF  MATERIALS. 

Slag. 


25 

62 

115 

11 


132 


Charges. 

. 

Steel. 

Metal, 

6,445  kg. 

c, 

58  " 

6 

Si, 

27  " 

2 

Mn, 

72  " 

10 

Fe, 

5,288  " 

5,173 

Limestone, 

20  " 

CaO, 

11  " 

.... 

c, 

2.5  " 

.... 

o, 

6.5  " 

.... 

Hearth, 

SiO2 

132  " 

.... 

Gas, 

7,884   " 

c, 

933   " 

.... 

0, 

2,003   " 

.... 

H, 

118  " 

.... 

N, 

4,830  " 

.... 

Air, 

16,026  " 

0, 

3,812  " 

.... 

N, 

12,195  " 

.... 

H, 

19  " 



Totals, 


29,507   " 


5,191 


80 


425 


Gases. 
52 


2.5 
6.5 


933 

2,003 

118 

4,830 


23,891 


NOTES  ON  THE  BALANCE  SHEET. 

The  distribution  of  carbon,  silicon,  manganese  and  iron  is 
governed  by  the  known  amounts  of  these  elements  present  in 
the  steel,  the  rest  of  the  carbon  going  into  the  gases  (as  CO2), 
and  the  manganese,  silicon  and  iron  passing  into  the  slag  (as 
MnO,  SiO2  and  FeO,  respectively.) 

The  amount  of  limestone  used  was  not  stated,  but  was  de- 
duced from  the  fact  that  the  slag  was  said  to  weight  425  kilos, 
and  to  contain  2.54  per  cent,  of  CaO,  which  makes  the  CaO 
11  kilos,  and  assuming  pure  limestone,  this  would  bring  in  9 
kilos,  of  CO2,  which  appears  on  the  balance  sheet  as  2.5  kilos, 
of  carbon  and  6.5  kilos,  of  oxygen,  contributed  to  the  gases. 


406  METALLURGICAL  CALCULATIONS. 

The  weight  of  silica  contributed  by  the  hearth  is  deduced 
from  the  fact  that  the  slag  must  contain  the  silicon,  manganese 
and  iron  oxidized  from  the  charge,  as  SiO2,  MnO  and  FeO,the 
CaO  of  the  flux  and  the  SiO2  contributed  by  the  hearth,  and  its 
total  weight  is  425  kilos.  The  ingredients  of  the  slag  must, 
therefore,  be 

Kg. 

SiO2  25X60/28  =  53.5 
MnO  62X71/55=  80.0 
FeO  115X72/56  =  147.9 
CaO  =  11.0 

.    Sum    293 
From  hearth  (difference)  132 

Total  slag  425 

The  gas  used  we  find  by  starting  with  the  fact  that  1,980 
kilos,  of  coal  is  used,  from  which  47.13  per  cent,  of  carbon 
enters  the  gases.  This  makes  carbon  in  the  gases  933  kilos. 
Each  cubic  meter  of  dry  gas,  as  analyzed,  contains  0.2805 
cubic  meter  of  CO2,  CO  and  CH4  added  together,  and  there- 
fore, 0.2805X0.54  =  0.1515  kilos,  of  carbon.  The  volume  of 
dry  producer  gas  used  is  therefore,  at  standard  conditions, 
933-^0.1515  =  6,160  cubic  meters,  containing  by  its  analysis: 

CO2  6,160X0.0381  =     235  m3  =     465  kg. 

O2  "     X  0.0098  =       60m3  =       86" 

CO  "     X  0.2382  =  1,467  m3  =  1,840   " 

CH4  "     X  0.0042  =        26m3=        19" 

H2  "     X  0.0875  =      539  m3  =        49" 

N2  "     X  0.6222  =  3,833  m3  =  4,830   " 


6,160  m3  =  7,297 

The  moisture  is  82  grams  per  each  cubic  meter  of  gas,  meas- 
ured at  26°  and  717  m.m.  pressure;  but  the  6,160  cubic  meters 
of  gas  at  standard  conditions  would  be  7,175  cubic  meters  at 
those  conditions  of  temperature  and  pressure,  and  therefore 
be  accompanied  by 

7,175X82  =  588,350  grams 

=  588  kg.  of  H2O  vapor. 


EFFICIENCY  OF  OPEN -HEARTH  FURNACES.  407 

We  can  now  enter  the  gas  on  the  balance  sheet  either  as  so 
much  CO2,  CO,  H2O,  etc.,  or  else  resolve  it  into  its  essential 
constituents  C,  H,  O  and  N,  which  course  we  have  followed 
on  the  balance  sheet.  The  carbon  in  the  gas,  is  933  kg.  by 
assumption, 'the  oxygen  is  8/11  the  CO2,  all  the  O2,  4/7  the  CO 
and  8/9  the  H2O,  a  total  amounting  to  2,003  kilos;  the  hydro- 
gen is  4/16  the  CH4,  all  the  H2  and  1/9  the  H2O,  a  total  of 
118  kilos. 

The  air  supplied  is  best  found  from  the  volume  of  the  chim- 
ney gases.  The  total  carbon  entering  these  is  52  +  2.5  +  933 
=  987.5  kilos.,  as  seen  from  the  balance  sheet.  Each  cubic 
meter  of  dry  chimney  gas  contains  0.1112  m3  of  CO2,  carrying 
0.1112X0.54  =  0.0600  kilos,  of  carbon.  The  volume  of  dry 
chimney  gas  at  standard  conditions  is  therefore  987.5-^0.0600 
=  16,458  m3.  This  contains  16,458X0.8210  =  13,512  m3  of 
N2,  and  since  3,833  cubic  meters  came  in  with  the  gas  9,679 
m3  must  have  come  in  with  the  air,  corresponding  to  12,220  m3 
of  dry  air  at  standard  conditions.  This  would  consist  of  12,195 
kilos,  of  N2  and  3,660  kilos,  of  O2.  To  find  the  moisture  present 
the  volume  of  this  air  at  26°  and  717  m.m.  pressure  would  be 
14,230  m3,  and  would  be  therefore  accompanied  by 

14,230X12  =  170,760  grams 
=  171  kg.  of  H2O. 

This  consists  of  19  kilos,  of  hydrogen  and  152  kilos,  of  oxygen, 
the  latter  increasing  the  total  oxygen  in  the  air  used  to  3  660  + 
152  =  3,812  kilos. 

(2)  HEAT  BALANCE  SHEET. 

Heat  Available. 

Per- 
Cal.     centages. 

Heat  in  the  warm  charges 189,210  =    2.45 

Sensible  heat  of  air  used 99,480  =    1 . 29 

Sensible  heat  of  gas  used 360,550  =    4.68 

Heat  of  combustion  of  gas .6,202,300  =  80.44 

Heat  of  oxidation  of  the  bath 833,600  =  10. 81 

Heat  of  formation  of  slag 24,200  =    0.31 

Total. .  7,709,340  =  100.00 


408  METALLURGICAL  CALCULATIONS. 

Heat  Distribution. 

In  melted  steel  at  tapping 1,437,900  =  18.65 

Decomposition  of  limestone 9,200  =  0. 12 

Sensible  heat  of  slag 238,000  =  3.09 

Sensible  heat  of  chimney  gases 3,118,450  =  40.45 

All  other  losses,  not  classified 2,905,790  =  37.69 

Total 7,709,340  =  100.00 

Notes  on  the  Heat  Balance  Sheet. 

The  warmed  charges  weighed  1,700  kilos,  at  700°,  and  the 
cold  charges  3,745  kilos  at  26°.  Taking  0°  C.  as  the  base  line 
the  sensible  heat  in  these  is 

Cat. 

1,700X0.15X700  =  178,500 
3,745X0.11  X   26  =     10,710 


Sum  =  189,210 

The  air  used  contains,  at  standard  conditions,  12,220  m3  of 
air  and  171  -f-  0.81  =  211  m3  of  water  vapor.  These  carry  at 
26°,  heat  as  follows: 

Col. 

12,220X0.3037X26  =  98,490 
211X0.3439X26  =        990 

Sum  =  99,480 
The  gas  used,  coming  in  at  165°  C.,  carries  in  heat  as  follows: 

Col. 

O2,  CO,  H2,  N2  5,899  m3X  0.3075  =  1,814 
CO2  235  m3X  0.4063  =  95 
CH4  26  m3X 0.4163  =  11 
H2O  726  m3X  0.3648  =  265 


Average  calorific  capacity  per  1°  =     2,185 
Heat  content  2, 185  X 165  =  360,525 

The  heat  of  combustion  is  that  of  the  combustible  ingredi- 
ents of  the  gas  used  to  CO2  and  H2O  vapor: 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  409 

Col. 

CO  1,467  m3X  3,062  =  4,492,000 
CH4  26  m3X  8,598  =  223,500 
H2  539  m3X  2,613  -  1  486,800 


Sum  =  6,202,300 

The  heat  of  oxidation  of  the  bath  is  from  the  various  sub- 
stances oxidized: 

Cal. 

C  to  CO2  52X8,100  =  421,200 

Si  to  SiO2  25X7,000  -  175,000 
Mn  to  MnO  62x1,653  =  102,500 
Fe  to  FeO  115X1,173  =  134,900 

Sum  =  833,600 

The  heat  of  formation  of  the  slag  is  the  heat  of  combination 
of  80  kilos,  of  MnO,  148  kilos,  of  FeO  and  11  kilos,  of  CaO, 
with  186  kilos,  of  SiO2.  This  will  be,  since  the  bases  are  largely 
in  excess  of  the  silica: 

186X130  =  24,200  Cal. 

The  figure  130  is  an  average  for  the  heat  of  combination  of 
1  kilo,  of  SiO2  with  about  2  parts  of  FeO  to  1  part  MnO. 

The  heat  in  the  steel  at  tapping  is  simply  its  weight  multi- 
plied by  its  heat  contents  per  kilo.  (5,191x277). 

The  heat  in  the  slag  is  similarly  obtained  (425x560). 

The  decomposition  of  limestone  represents  9  kilos,  of  CO3 
liberated,  and  the  heat  absorbed  is  9x1,026. 

The  sensible  heat  in  the  chimney  gases  is  obtained  by  first 
noting  that  their  volume  (measured  dry),  as  already  obtained, 
is  16,458  cubic  meters.  The  CO2,  11.12  per  cent.,  becomes, 
therefore,  1,843  m3;  the  O2,  1,116  m3;  N2,  13,512  m3,  while  the 
H2O  accompanying  this  will  be  9  times  the  weight  of  hydrogen 
passing  into  the  gases,  which  is  9X  (118+17)  =  1,215  kilos. 
=  1,500  m3.  A  simpler  way  to  get  the  volume  of  the  water 
vapor  is  to  observe  that  it  is  always  equal  to  the  volume  of 
hydrogen  going  into  it,  and,  therefore,  in  this  case  would  be 
(1 18+ 17) -T- 0.09  =  1,500  ms.  The  heat  carried  out  by  these 
gases  would  therefore  be: 


410  METALLURGICAL  CALCULATIONS. 

N2  +  02   =  14,628  m3X 0.3165  =     4629.8  Cal.  per  V 
CO2  =     1, 843  m3X 0.4800  =       984.4     " 
H20  =     1,500  m3X 0.4150  =       622.5     " 


Calorific  capacity  =     6236.9     " 

Total  capacity  =  3,118,450    "  per  500° 

The  heat  balance  sheet  as  a  whole  discloses  the  fact  that  in 
this  furnace  the  fuel  only  supplies  some  80.5  per  cent,  of  the 
total  heat  available,  and  that  about  10.8  per  cent,  is  furnished 
by  the  oxidation  of  the  bath  itself.  On  the  other  hand,  the 
melted  steel  accounts  for  18.6  per  cent.,  while  chemical  reactions 
absorb  almost  none,  giving  a  net  thermal  efficiency  of  slightly 
under  19  per  cent.  The  other  important  items  are  40.5  per  cent, 
of  the  total  heat  lost  up  the  chimney,  and  38  per  cent,  lost  by 
radiation  and  conduction.  Such  data  as  these  point  the  way 
to  avenues  of  possible  saving  or  avoidance  of  waste  of  heat. 

(3)  The  excess  of  air  is  obtained  directly  from  the  chimney 
gases.  The  1,116  m3  of  oxygen,  unused,  represents  1,116-s- 
20.8  =  5,365  m3  of  air  in  excess,  which  leaves  16,458—5,365 
=  11,093  m3  of  air  which  came  in  and  was  used.  The  per- 
centage of  air  used  in  excess  of  that  which  was  necessary  was: 

5,365-4-11,093  =  0.4845  =  48.5  per  cent.  (3) 

No  properly  run  open-hearth  regenerative  gas  furnace  should 
ever  have  such  a  large  excess  of  air,  for  it  cuts  down  the  tem- 
perature of  the  flame  and  leads  to  high  chimney  losses. 

The  air  used  brought  in  171  kilos  of  water  vapor,  and  there- 
fore the  excess  air  brought  in 

48  5 

171X7I^-H  =  56  kilos-  of  water» 
14o.  O 

the  volume  of  which,  at  standard  conditions,  would  be 
56^-0.81  =  61  cubic  meters. 

The  excess  air,  with  its  accompanying  water,  going  into  the 
chimney  at  500°,  carried  out  heat  as  follows: 

Cal. 

5,365X0.3165X500  =  849,000 
61X0.4150X500  =     12,650 


861,650 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  411 

Representing  861,650-5-7,709,340  =  0.112  =  11.2  per  cent.      (3) 
(4)  The  thermal  efficiency  has  been  already  added  up  as 

18.65  +  0.12  =  18.77  per  cent.  (4) 

Problem  73. 

In  the  open-hearth  furnace  of  the  preceding  problem, 
assume  that  the  calculations  therein  made  showed  that,  per 
heat  of  steel  produced,  5,191  kilograms,  there  entered  and 
left  the  furnace  the  following  volumes  of  gases,  measured  at 
standard  conditions: 

Producer  Gas.                 Air.  Chimney  Gases. 

CO2  235m3             1,830m3 

O2  60m3  3,833m3  1,116m3 

CO  1,467m3             

CH4  26m3             

H2  539m3            

N2  3,834m3  9,679m3  13,512m3 

H2O  726m3             211m3  1,500m3 

The  temperature  of  air  used  was  26°,  of  producer  gas  165°, 
of  chimney  gases  400°.  The  excess  of  air  used  was  48.5  per 
cent.  The  gas  and  air  entered  the  laboratory  of  the  furnace 
preheated  to  1,100°,  and  the  products  of  combustion  entered 
the  regenerators  at  1,450.  Items  of  heat  balance  sheet: 

Available. 

Calories 

In  warm  charges 189,210 

Sensible  heat  of  air  used  at  26° 90,480 

Sensible  heat  of  gas  used  at  165° 360,550 

Heat  of  combustion  of  the  gas 6,202,300 

Heat  of  oxidation  of  the  bath 833,600 

Heat  of  formation  of  slag 24,200 


7,700,340 
Distribution. 

In  melted  steel  at  tapping  at  1,410° 1,437,900 

In  slag  at  tapping  at  1,410° 238,000 

Decomposition  of  limestone 9,200 

Sensible  heat  in  chimney  gases  at  400° 3,065,350 

All  other  losses,  not  classified . .  .  .  2,949,890 


7,700,340 


412  METALLURGICAL  CALCULATIONS. 

Required : 

(1)  The  thermal  efficiency  of  the  regenerators. 

(2)  The  thermal  efficiency  of  the  laboratory  of  the   furnace. 

(3)  The  temperature  of  the  flame. 

.  (4)  The  change  in  (3)  if  only  the  theoretical  amount  of  air  for 
combustion  were  used. 
Solution : 

(1)  The  products  of  combustion,  entering  the  regenerators 
at  1,450°,  carry  into  them  the  following  amounts  of  heat: 

Calories. 

CO2  1,830X0.689  =  1,261 

O2  +  N2  14,628X0.342  =  5,003 

H2O  1,500X0.557  =     836 

Calorific  capacity  per  1°  =  7,100 

Calorific  capacity  per  1,450°  =  10,295,000  Calories. 

The  same  gases  entering  the  chimney  flue  at   400°  carry 
with  them  the  following  amounts : 

Calories. 

CO2  1,830X0.458  =     838.1 

O2  +  N2  14,628X0.314  =  4,593.2 

H20  1,500X0.400  =     600.0 


Calorific  capacity  per  1°  =  6,031.3 

Calorific  capacity  per  400°  =  2,412,500  Calories. 

The  gas  used  for  combustion,  entering  the  regenerators  at 
165°  and  leaving  it  at  1,100°,  carried  into  the  regenerators 
360,550  Calories,  as  already  given  in  the  balance  sheet,  and 
carried  out  at  1,100°  the  following: 

Calories. 

CO2  235X0.612  =      143.8 

CH4  26X0.620  =        16.1 

O2,  N2,  H2,  CO  5,899X0.333  =  1,964.4 

H2O  726X0.505  =      366.6 


Calorific  capacity  per  1°  =  2,490.9 

Calorific  capacity  per  1,100°  =  2,780,000  Calories. 
Heat  abstracted  from  regenerators: 

2,780,000—360,550  =  2,419,450  Calories. 


EFFICIENCY  OF  OPEN-HEAR!  H  FURNACES.  413 

The  air  used,  entering  the  regenerators  at  26°,  carries  in  as 
sensible  heat  90,480  Calories,  as  already  given  in  the  balance 
sheet,  and  issuing  from  them  at  1,100°  carries  out  the  fol- 
lowing : 

Calories. 

02  +  N2  13,512X0.333  =  4,499.5 

H2O  211X0.405=        85.5 


Calorific  capacity  per  1°  =  4,585.0 

Calorific  capacity  per  1,100°  =  5,043,500  Calories. 
Heat  abstracted  from  regenerators  : 

5,043,500—90,500  =  4,953,000  Calories. 

The  thermal  efficiency  of  the  regenerators  may  now  be  cal- 
culated from  three  standpoints.  There  is  no  doubt  that  the 
gas  and  air  take  from  the  regenerators,  and  return  to  the  body 
of  the  furnace  2,419,450  +  4,953,000  =  7,372,450  Calories.  This 
is,  therefore,  the  usefully  returned  heat,  and  the  ratio  of  this  to 
the  heat  received  by  the  regenerators  measures  their  efficiency 
qua  regenerators.  The  three  figures  obtained  for  this  efficiency 
depend  on  what  is  to  be  considered  as  the  heat  chargeable 
against  the  regenerators.  Are  they  to  be  charged  with  all  the 
heat  in  the  hot  products  at  1,450°  (10,295,000  Calories),  or  only 
with  the  heat  left  in  the  regenerators  by  these  products  leaving 
at  400°  (10,295,000—2,412,500  =  7,882,500  Calories),  or  per- 
haps only  with  the  heat  carried  in  less  a  certain  assumed  amount 
representing  the  minimum  temperature  to  which  it  is  desirable 
to  cool  the  products  before  they  enter  the  chimney? 

If  we  charge  the  regenerators  with  all  the  heat  brought  in 
by  the  products  their  thermal  efficiency  figures  out: 


=  0.72  =72  per  cent.  (!) 


If  we  charge  them  with  the  heat  left  in  the  regenerators,  by 
the  products,  their  efficiency  is: 

7,372,450 

75^600  =  0-94  =94  per  cent.  (1) 


414  METALLURGICAL  CALCULATIONS. 

leaving  7  per  cent,  of  the  heat  chargeable  against  them  lost  by 
radiation  from  their  walls  and  conduction  to  the  ground. 

If  we  think  that  the  first  calculation  gives  too  low  an  effi- 
ciency, because  the  gases  must  leave  the  regenerators  hot,  in 
order  to  be  used  for  chimney  draft,  and  therefore  some  or  all 
of  the  heat  in  the  chimney  gases  should  not  be  charged  against 
the  regenerators,  on  the  other  hand,  the  second  calculation  may 
represent  too  high  an  efficiency,  because  the  gases  may  be  dis- 
carded to  the  chimney  at  a  higher  temperature  than  is  neces- 
sary to  provide  the  requisite  chimney  draft,  and  this  excess  of 
chimney  temperature  and  consequent  heat  loss  is  a  defect  of 
the  regenerators  which  they  should  be  charged  with.  If  we 
assume  that  a  chimney  gives  very  nearly  its  maximum  drawing 
capacity  with  the  gases  entering  it  at  300°,  it  would  be  per- 
fectly proper  to  charge  the  regenerators  with  all  the  heat  which 
could  be  given  out  by  the  products  in  cooling  from  1,450° 
to  300°,  heat  which  they  should  have  entirely  absorbed.  In 
the  case  in  hand,  the  products  at  300°  would  contain  (by  calcu- 
lations similar  to  those  already  made)  1,777,400  Calories,  leaving 
10,295,000—1,777,400  =  8,517,600  Calories  chargeable  against 
the  regenerators,  as  the  heat  which  they  should  have  absorbed 
or  intercepted.  Measured  by  this  standard,  their  efficiency  is: 


The  losses  from  the  regenerators  in  this  view  would  be  7  per 
cent,  (about)  by  radiation  and  7  per  cent,  by  unnecessary  chim- 
ney loss. 

Comparing  these  three  methods  of  considering  efficiency, 
the  third  appears  to  the  writer  as  the  fairest,  and  the  one  which 
gives  the  metallurgist  the  most  reliable  criterion  of  the  real 
work  which  his  regenerators  are  doing  for  him,  and  the  best 
basis  of  comparison  when  considering  the  work  of  different 
regenerators  or  of  the  same  regenerators  under  different  con- 
ditions. 

(2)  The  laboratory  of  the  furnace,  the  space  enclosed  be- 
tween the  hearth,  roof,  side  walls  and  ports,  receives  from  the 
entering  preheated  gas  and  air  their  sensible  heat,  and  the  heat 
generated  by  combustion.  In  the  case  in  point  these  items 
total,  as  already  calculated: 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.          415 

Calories. 
Sensible  heat  in  preheated  gas  .........   2,722,300 

Sensible  heat  in  preheated  air  ..........   5,043,500 

Heat  generated  by  the  combustion  .....   6,202,300 

Total  ...........................  13,968,100 

Hot  products  at  1,450°  take  out  .......  10,295,000 

Heat  left  in  the  laboratory  ............   3,673,100 

These  figures  show  that  the  laboratory  of  the  furnace  ap- 
propriates to  its  own  purposes  3,673,100  Calories  out  of  the 
13,968,100  Calories  poured  into  it.  This  part  of  the  furnace, 
therefore,  qua  laboratory,  has  an  efficiency  in  this  respect  of 


(2) 


This  is  the  datum  which  would  be  useful  to  the  metallurgist 
in  comparing  the  efficiencies  of  differently  shaped  laboratories, 
such  as  those  with  differently  shaped  hearths,  differently  shaped 
roof,  differently  arranged  ports,  etc.  This  conception  of  effi- 
ciency is  that  taken  by  Damour,  and  repeated  by  Queneau 
in  his  book  on  "  Industrial  Furnaces."  We  must  be  careful 
here  not  to  compare  the  heat  left  in  the  laboratory  with  the 
heat  of  combustion  alone.  This  would  give 

3,673,100 

-  °'59  -  59 


But  this  is  not  the  heat  absorption  of  the  laboratory  alone,  but 
is  a  function  of  the  furnace  as  a  whole,  and  depends  largely  on 
the  efficiency  of  the  regeneration  accomplished  by  the  regen- 
erators. If  we  wish  to  obtain  an  idea  of  the  perfection  with 
which  the  laboratory  of  the  furnace  appropriates  the  heat  pass- 
ing into  and  through  it,  we  must  simply  compare  what  it  ap- 
propriates with  what  was  sent  into  it.  The  percentage  of  this 
appropriation  measures  the  efficiency  of  the  laboratory  for 
abstracting  heat  for  the  purpose  of  heating  itself  —  a  datum 
highly  useful  to  know  if  the  furnace  is  used  simply  for  keeping 
a  given  working  space  up  to  a  given  temperature  for  an  in- 


416  METALLURGICAL  CALCULATIONS. 

definite  time.  If  the  heating  of  the  furnace  charge  to  the  furnace 
heat  is  only  a  minor  part  of  the  useful  work  of  the  furnace, 
and  keeping  it  at  that  temperature  is  the  chief  function  of  the 
furnace,  then  the  relation  of  the  heat  thus  appropriated  and 
utilized  to  the  total  heat  available  to  the  laboratory,  is  a  measure 
of  the  perfection  of  construction  of  the  laboratory.  This 
is  the  view  of  Damour  and  Queneau,  and  seems  in  some  re- 
spects plausible. 

However,  if  we  are  examining  this  question  thus  in  detail, 
it  appears  to  the  writer  that  the  view  just  explained  and  the 
conclusions  derived  thereform  may  really  be  the  very  opposite 
of  the  truth  of  the  matter,  and  lead  to  entirely  erroneous  con- 
clusions. If  two  exactly  similar  open-hearth  furnaces  are 
constructed,  with  the  sole  difference  that  the  body  of  one  is 
thicker  walled  than  the  body  of  the  other,  it  is  perfectly  true 
that  maintaining  the  same  temperature  in  each  will  require 
more  gas  in  the  thin- walled  furnace,  but  that  it  will  abstract 
and  radiate  a  considerably  larger  proportion  of  the  heat  passing 
through  it  than  the  thick- walled  furnace.  Yet,  according  to 
Damour  and  Queneau's  principle,  the  thin-walled  furnace 
would  be  considered  as  being  the  more  efficient  laboratory  of 
the  two.  The  truth  is,  we  must  either  compute  the  heat  left 
in  the  -laboratories  per  unit  of  time  per  cubic  meter  of  stock 
space  in  order  to  compare  different  furnaces,  or  else  to  get 
absolute  efficiency  compute  the  ratio  of  the  heat  passing  into 
the  laboratory  with  that  absorbed  usefully  by  the  charge,  viz.: 
in  this  case: 


1,447,100       A  1A/1       1A  A 

0.104  =  10.4  per  cent. 


13,968,100 


(2) 


(3)  The  temperature  of  the  flame  is  that  to  which  the  pro- 
ducts of  combustion  can  be  raised  by  the  maximum  heat  which 
they  contain.  When  perfectly  consumed  they  contain  the  heat 
pre-existing  as  sensible  heat  in  the  hot  gas  and  air,  plus  the 
heat  of  combustion.  We  have  already  calculated  this  total  as 
13,968,100  Calories.  The  mean  heat  capacity  of  the  products 
of  combustion,  per  1°  between  0°  and  t°,  is: 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.         417 

CO2  1,830  m3  (0.37   +0.0002210     =     677  +  0.4026t 

O2  +  N2   14,628  ms  (0.303  +  0.000027t)  =  4,428  +  0.3950t 
H2O          1,500  m3  (0.34   +0.00015t)     =      510  +  0.2250t 


Sum  =  5,615  +  1.0226t 
Therefore, 

13,968,100 
~  5,615  +1.0226t 

Whence, 

t  =  1,749°  (3) 

(4)  If  the  excess  air  were  not  used,  the  1,116  cubic  meters  of 
oxygen  in  the  products,  together  with  its  corresponding  quan- 
tity of  nitrogen,  4,249  cubic  meters,  would  be  absent  from  the 
products  of  combustion,  as  would  also  some  of  the  211  cubic 
meters  of  water  vapor.  Since  the  oxygen  in  the  air  was  3,833 
m3  and  the  unused  excess  1,116,  the  proportion  of  the  211  m3 
of  water  vapor  belonging  to  the  excess  air  was 


and  the  final  products  are  CO2   1,830    m3,   N2    9,263  m3    and 
H20  1,439  m3. 

The  heat  available  will  be  the  same  as  before  from  com- 
bustion, the  same  from  preheated  gas,  but  less  from  preheated 
air,  because  of  the  absence  of  the  excess  air.  Since  the  air 
formerly  used  brought  in  5,043,500,  the  proportion  of  this 
carried  by  the  excess  air  is 

5,043,500  X    L-=  1,468,500, 


leaving  heat  brought  in  by  air  in  the  second  case  the  difference, 
viz.:  3,575,000,  and  the  total  heat  in  the  products  13,968,100— 
1,468,500  =  12,499,600  Calories. 

The  mean  heat  capacity  of  the  diminished  quantity  of  pro- 
ducts, per  1°,  is,  by  the  same  methods  as  previously  used, 
3,973  +  0.8685t,  and,  therefore, 

12,499,600 


3,973 +  0.8685t 


418  METALLURGICAL  CALCULATIONS. 

a  gain  of  nearly  400°  in  maximum  temperature,  by  cutting  off 
the  40  per  cent,  excess  of  air  which  was  being  used. 

Problem  74. 

The  new  style  Siemens  furnace  for  small  plants  has  the  gas 
producers  built  in  as  part  of  the  furnace,  between  the  two  re- 
generators. The  furnace  has  only  one  regenerative  chamber 
at  each  end  for  preheating  the  air  used  for  burning  the  gas, 
while  the  latter  comes  to  the  ports  hot  directly  from  the  pro- 
ducers. This  plant  is  compact,  economical  of  fuel,  allows  high 
temperatures  to  be  reached,  and  occupies  little  floor  space.  As 
compared  with  the  old  style  separate  furnace  and  producer 
plant,  it  occupies  about  half  the  floor  space,  costs  about  60 
per  cent,  as  much  to  build  and  uses  about  60  per  cent,  as  much 
coal.  They  are  largely  used  abroad  for  melting  steel  for  cast- 
ings in  small  foundries  and  for  melting  pig  iron  for  castings  to 
be  subsequently  annealed  to  malleable  castings. 

Such  a  furnace  melted  a  charge  of  3,000  kilograms  of  cast 
iron  in  4  hours,  using  750  kilograms  of  coal.  The  cast  iron 
was  charged  cold,  at  0°,  and  run  out  melted  at  1,450°,  con- 
taining 300  Calories  of  sensible  heat  per  kilogram.  The  coal 
and  gases  were  of  the  following  compositions: 

Coal— C  75,  H  5,  O  12,  H2O  2;  ash  6. 

Producer  Gas— CO  20,  CO2  5,  CH4  2,  H2  16,  N2  57. 

Chimney  Gas— CO2 19,  O2 1.8,  N2  79.2. 

The  air  coming  to  the  furnace  is  at  0°  and  dry ;  a  steam  blower 
is  used  to  run  the  producer,  using  1  kilo,  of  steam  per  6  kilos, 
of  air  blown  in,  and  the  raising  of  this  steam  requires  0.1  kilo 
of  coal  used  under  boilers.  The  ashes  produced  weigh  75  kilos, 
per  heat  run.  Temperature  of  the  hot  producer  gas  entering 
the  furnace  laboratory  600°,  of  preheated  air  1,000°,  of  pro- 
ducts leaving  laboratory  1,400°,  of  products  entering  chimney 
flue  350°. 
Required : 

(1)  A  heat  balance  sheet  of  the  furnace  as  a  whole. 

(2)  The  net  thermal  efficiency  of  the  furnace. 

(3)  The  net  thermal  efficiency  of  the  laboratory  of  the  fur- 
nace. 

(4)  The  net  thermal  efficiency  of  the  regenerators. 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.  419 

(5)  The  theoretical  flame  temperature  in  the  furnace. 
Solution : 

(1)   Heat  balance  sheet  per  charge  of  3,000  kilos. 

Heat  Available. 

Calories. 

Calorific  power  of  coal  used  in  producers 5,248,500 

<f       "     "        "     under  boilers '.  .  .     398,900 


5,647,400 
Heat  Distribution. 

In  melted  cast  iron 900,000 

In  chimney  products 768,950 

Loss  by  unburnt  carbon  in  ashes 243,000 

Used  for  raising  steam 398,900 

Loss  by  radiation  and  conduction 3,336,550 


5,647,400 

.     Calculation  of  the  Heat  Balance. 

The  heat  of  combustion  of  a  kilogram  of  coal  is  not  given, 
therefore  must  be  calculated  from  its  composition: 

Calories. 
CO.  75X8,100  6,075 

X  34,500  =          1,208 


Calorific  power  to  liquid  water  =          7,283 

Vaporization  heat  of  water  formed 

(0.02  + 0.45)  X  606.5  =  285 


Calorific  power  to  vapor  of  water  6,998 

Of  750  kilos,  in  the  producers  =  5,248,500 

The  coal  used  under  the  boilers  can  only  be  found  by  first 
finding  how  much  steam  was  used,  which  in  its  turn  can  be 
gotten  from  the  air  blown  in,  and  the  nitrogen  of  this  can  be 
found  from  the  total  nitrogen  in  the  producer  gas.  The  volume 
of  producer  gas  can  be  gotten  from  its  carbon  content  per  cubic 
meter  and  the  known  weight  of  carbon  gasified.  Or,  turning 
this  chain  of  reasoning  the  other  way,  if  we  subtract  the  carbon 
in  the  ashes  from  the  carbon  in  the  coal,  the  difference  is  the 
carbon  entering  the  gases;  this  divided  by  the  carbon  in  1 


420  METALLURGICAL  CALCULATIONS. 

cubic  meter  of  chimney  gas  (calculated  from  its  analysis)  gives 
the  volume  of  chimney  gas,  and  by  the  carbon  in  1  cubic  meter 
of  producer  gas  gives  the  volume  of  this  gas  used,  from  which 
the  nitrogen  in  it  can  be  calculated,  which  divided  by  0.792 
gives  the  volume  of  air  used,  from  which  its  weight  is  obtained, 
thence  the  amount  of  steam  used,  and  finally  from  this  the 
amount  of  coal  necessary  to  burn  under  boilers  to  raise  this 
steam. 

[In  working  most  metallurgical  problems  the  difficulty  of 
finding  the  connection  or  succession  of  steps  connecting  the 
requirements  with  the  data  given  is  often  easiest  overcome  by 
starting  with  the  requirement  in  mind  and  noting  from  what 
other  figure  its  value  may  be  calculated,  and  thus  passing 
backwards  from  one  figure  to  another  we  finally  arrive  at  one 
which  can  be  found  directly  from  the  data  given.  The  logical 
sequence  of  operations  is  thus  disclosed  and  the  problem  is  in 
reality  solved;  the  following  of  the  thread  in  the  reverse  direc- 
tion, from  data  to  requirement,  involves  calculations  only,  not 
hard  thinking,  and  is  usually  a  matter  of  the  simplest  arith- 
metic.] 

The  operations  for  calculating  the  steam  used  are  as  follows: 

Carbon  in  coal  in  producers  =  750x0.75  =      562.5  kg. 

Carbon  in  ashes  from  producers  =  75 — 45  =        30.0   " 


Carbon  in  gases  from  producers  =  532.5   " 

Carbon  in  producer  gas  per  1  m3  =  0.27x0.54  =  0.1458  " 

Producer  gas  per  heat  =  532.5-^0.1458  =  3,652m3 

Nitrogen  in  producer  gas  =  3,652x0.57  =  2,082   " 

Air  used  in  the  producer  =  2,082-^0.792  =  2,629  " 

=  3,399  kg. 

Carbon  in  chimney  gas  per  1  m3  =  0.19x0.54  =  0.1026   " 

Chimney  gas  per  heat  =  532.5-^0.1026  =  5,190  m3 

Nitrogen  in  chimney  gas  =  5,190X0.792  =  4,110   " 

Nitrogen  from  air  used  =  4,110—2,082  =  2,028  " 

Air  used  to  burn  gas  =  2,028-^0.792  =  2,561   " 

Weight  of  this  air  =  2,561X1.293  =  3,311   " 

Weight  of  air  used  in  producer  =      3,399   " 

Weight  of  steam  used  in  producer  =  3,399-^6  =  567   " 

Weight  of  boiler  coal  used  =  567X0.1  57   " 

Calorific  power  of  this  coal  =  6,998X57  =  398,900  Cal, 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.          421 

The  heat  in  the  melted  cast  iron  is  simply  its  weight  times 
300  =  300X3,000  =  900,000  Calories. 

(2)  The  net  thermal  efficiency  follows  directly  from  this,  as 

900,000  0.16  =  16  per  cent.  (2) 


5,647,400. 

(3)  The  gaseous  products  consist  dry,  as  analyzed,  of 

CO2       5,190X0.19    =      986m3 
O2         5,190X0.018  =        93   " 
N2         5,190X0.792  =  4,110  " 

And  in  addition  all  the  water  formed  from  the  coal,  plus  the 
steam  used,  a  total  of: 

Water  from  coal  =  0.47X750  =  352.5  kg. 

Steam  =  566.5   " 


Total  water  vapor  in  gases  =919.0   " 

Volume  at  standard  conditions  =  919-^-0.81  =  1,135m* 

At  350°  these  products  carry  out  of  the  furnace  heat  as  fol- 
lows: 

Calories. 

O2  +  N3  4,203X0.312  =  1,311 

CO2  986X0.447  =      441 

H2O  1,135X0.392  =      445 

Average  caloric  capacity  per  1°  =  2,197 

Capacity  per  350°  =  768,950  Calories. 

These  same  products  would  carry  heat  out  of  the  laboratory 
of  the  furnace  at  1,400°  as  follows: 

Calories. 

O2  +  N2  4,203X0.341  =  1,433 

CO2  986X0.678  =     669 

H2O  1,135X0.550  =     624 


Average  caloric  capacity  per  1°  =  2,726 

Capacity  per  1,400°  =  3,816,400  Calories. 

The  heat  developed  in  the  laboratory  of  the  furnace  by  the 
combustion  of  the  3,652  m3  of  producer  gas  is 


422  METALLURGICAL  CALCULATIONS. 

Calories. 

CO  0.20X3062    =    612 

CH4  0.02X8,598  =      172 

H2  0.16X2,613  =      418 


Per  1m3  =  1,202 

Per  3,652  m3  =  4,389,700  Calories. 

The  3,652  m3  of  gas  comes  into  the  laboratory  of  the  furnace 
at  600°,  and  therefore  carries  in  as  sensible  heat  the  following: 

CO2  0 . 05  X  0 . 502  =  0 . 0251  Cal.  per  1° 

CH4  0.02X0.512  =  0.0102      " 

CO,  N2,  H2  0.93X0.319  =  0.2967      "       " 


0.3320X600  =199.2  Cal. 
Per  3,652  m3  =  199.2x3,652  =  727,480  Calories. 

This  does  not  count,  however,  the  water  vapor  accompanying 
the  producer  gas.  This  is  best  determined  from  the  fact  that 
the  total  hydrogen  in  the  coal  and  steam  used  in  the  producer 
must  appear  as  hydrogen  in  the  producer  gas  in  some  form  or 
other,  and  whatever  is  not  present  in  the  dry  producer  gas 
must  be  in  the  water  vapor  accompanying  it.  The  amount  of 
this  and  the  heat  it  carries  into  the  laboratory  of  the  furnace 
are  thus  determined: 

Hydrogen  in  750  kg.  coal  =  750X0.0522      =      39.15  kg. 
Hydrogen  in  steam  used  =  567-^-9  =      63.90   " 

Hydrogen  going  into  producer  gases  =     102.15   <: 

Hydrogen  in  dry  producer  gases, 

3,652X0.20X0.09=      65.74   " 
Hydrogen  in  producer  gas  as  vapor  of 

water  =      36.41   " 

Water  vapor  in  producer  gas  =       327.7   " 

Volume  of  this  vapor  405   " 

Heat  in  this  at  600°,  405X0.43X600  =  104,500  Cal. 

Total  heat  in  moist  producer  gas, 

727,480+104,500  =  831,980  " 

We  must  also  calculate  the  heat  brought  into  the  laboratory 
of  the  furnace  by  the  preheated  air,  at  1,000°,  used  for  com- 
bustion. This  is 

2,561  m3X 0.330 XI, 000  =  845,130  Calories. 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.          423 

It  follows  from  the  above  calculations  that  the  laboratory  of 
the  furnace  receives  per  heat  of  steel: 

Calories. 

From  hot  producer  gas  at  600°  =      831,980 

From  preheated  air  at  1,000°  =      845,130 

From  combustion  =  4,389,700 


Total  =  6",066,810 

Of  this  total  there  is  rejected  in  the  hot  products  at  1,400°, 
3,816,400  Calories,  leaving  in  the  laboratory  of  the  furnace 
2,250,410  Calories. 

According  to  the  view  of  Damour  and  Queneau  (Industrial 
Furnaces,  page  56)  the  ratio  of  the  heat  thus  left  in  the  body 
of  the  furnace  to  the  calorific  power  of  the  fuel  used,  measures 
the  thermal  efficiency  of  the  furnace.  According  to  that  view 
this  new  style  Siemen's  furnace  has  an  efficiency  of 

=  0.400  =  40.0  per  cent. 

This  figure,  however,  is  an  illusive  one.  It  enables  us  to 
compare  two  furnaces  and  see  which  regenerates  the  waste 
heat  best,  or  which  furnace  laboratory  is  designed  best  so  as 
to  catch  and  retain  most  of  the  heat  furnished  to  it;  but  the 
.real  question  is  the  comparison  of  different  laboratories  as  to 
their  net  melting  efficiency.  This  laboratory  abstracts  from 
the  heat  supply  furnished  to  it  2,250,410  Calories,  and  fur- 
nishes to  the  steel  900,000  Calories.  Its  proportion  of  usefully 
applied  heat  to  heat  appropriated  is,  therefore: 

=  0.400  =  40.0  per  cent.  (3) 

The  laboratory  of  the  furnace  therefore  loses  by  radiation 
60  per  cent,  of  the  heat  which  it  abstracts  from  the  gases,  or 
2,250,410-900,000  =  1,350,410  Calories.  This  is  24  per  cent, 
of  the  calorific  power  of  the  coal  used. 

'  (4)  The  regenerators  receive  3,816,400  Calories  from  the 
laboratory  of  the  furnace,  and  return  to  it  the  heat  in  the  pre- 
heated air  at  1,000°,  viz.:  845,130  Calories.  If  we  call  the 
ratio  of  these  two  the  efficiency  of  the  regenerators  we  have 

04  K  1  qrv 

AL  =  0.222  =  22.2  per  cent. 


424  METALLURGICAL  CALCULATIONS. 

If,  however,  we  charge  them  only  with  the  heat  actually  left 
in  them  by  the  hot  products,  entering  1,400°  and  leaving  at 
350°,  we  have  an  efficiency  of 


3°  °'277  "  *7'7  per  Cent' 


3.816,400-768,950 

As  long  as  the  chimney  gases  are  near  in  temperature  to  300°, 
we  can  use  the  latter  form  of  calculation  as  representing  the 
real  efficiency  of  the  regenerator.  The  regenerators,  there- 
fore, lose  by  radiation  and  conduction  to  the  air  2,202,320 
Calories,  which  is  72.3  per  cent,  of  the  heat  left  in  them  by  the 
hot  gases  and  39  per  cent,  of  the  calorific  power  of  the  coal 
used. 

(5)  The  flame  temperature  is  that  to  which  the  6,066,810 
Calories  available  in  the  laboratory  of  the  furnace  will  raise 
the  products  of  combustion.  The  average  mean  specific  heat 
of  the  latter  per  1°  is  : 

CO2  986  m3X  (0.37  +0.00022t)  =      365  +  0.2169t 

O2  +  N2   4,203  m3X  (0.303  +  0.0000271)  =  1,274  +  0.11351 
H2O         1,135  m3X  (0.34  +0.00015t)  =     386  +0.7031 

Sum  =2,025  +  0.50081 
6,066,810 


Therefore 


2,025+0.50081 
Whence  t  =  2,003  (5) 

Problem  75. 

In  an  basic  lined  open-hearth  furnace  using  the  Monell  pro- 
cess, 50  short  tons  of  melted  pig  iron,  at  1,300°,  is  run  upon 
30,000  pounds  of  Lake  Superior  iron  ore  (90  per  cent.  Fe2O3, 
10  per  cent.  SiO2)  lying  on  the  hearth  and  previously  heated  to 
1,300°.  There  is  2,000  pounds  of  burnt  lime  lying  on  the  ore 
to  help  to  form  slag.  The  ore  reacted  quickly,  almost  vio- 
lently, on  the  melted  pig  iron,  so  that  at  the  close  of  the  re- 
action, in  20  minutes,  the  slag  could  be  run  off.  The  pig  iron 
contained, 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.          425 

On  Running  in.  After  the  Reaction 

Carbon 3.50  3.00 

Silicon 2.00  0.00 

Phosphorus 0.75  0.00 

Manganese 0.50  0.00 

Iron 93.25  97.00 

The  oxidation  of  C,  Si,  P  and  Mn  may  be  assumed  as  being 
produced  first  by  reducing  all  Fe2O3  present  to  FeO,  and  com- 
pleted by  the  reduction  of  some  FeO  to  Fe.  The  carbon  is 
assumed  oxidized  in  the  reaction  to  CO,  although  this  CO  may 
be  subsequently  partly  burned  to  CO2  by  excess  air  in  the 
furnace. 
Required: 

(1)  The  amount  of  iron  reduced  into  the  bath. 

(2)  The    weight   and  percentage   composition     of    the   slag 
formed. 

(3)  The  items  of  heat  evolved  and  absorbed  in  the  reaction. 
Solution : 

(1)  We  cannot  make  a  simple,  direct  solution,  because  the 
weight  of  the  metal  after  the  reaction,  analysis  of  which  is 
given,  is  not  known.  The  inexperienced  calculator  might  be 
tempted  to  say  that  there  was  3.50 — 3.00  =  0.50  per  cent,  of 
carbon  oxidized,  2.00  per  cent,  of  silicon,  0.75  per  cent,  of  phos- 
phorus and  0.50  per  cent,  of  manganese,  calculate  the  weights 
of  these,  reckon  up  the  oxygen  they  would  absorb  in  being 
oxidized,  and  thus  get  at  the  amount  of  Fe2O3  reduced  to  FeO 
and  FeO  reduced  to  Fe.  This  would  be  correct  as  far  as  silicon, 
manganese  and  phosphorus  are  concerned,  because  they  are 
entirely  oxidized,  but  incorrect  for  the  weight  of  carbon,  be- 
cause the  3.00  per  cent,  not  oxidized  is  per  cent,  of  the  final 
bath,  which  is  of  different  weight  from  the  original  one,  and, 
therefore,  we  are  in  error  in  subtracting  3.00  from  3.50.  Being 
confronted  with  this  dilemma  we  can  see,  however,  that  if 
we  only  knew  the  weight  of  the  bath  after  the  reaction  we 
could  calculate  correctly  how  much  carbon  is  in  it,  thence  get 
the  correct  weight  of  carbon  oxidized,  then  the  correct  amount 
of  oxygen  absorbed  in  the  reaction,  and  from  this  the  weight 
of  iron  reduced  into  the  bath.  We  could  also  get  the  latter 
quantity  more  simply  by  finding  the  iron  present  in  the  bath 


426  METALLURGICAL  CALCULATIONS. 

after  the  reaction  (97  per  cent.),  and  subtracting  from  it  the 
iron  in  the  original  bath.  We  thus  have  two  ways  of  getting 
the  same  requirement  if  we  only  know  the  weight  of  the  bath. 
In  such  a  case  the  mathematical  key  to  the  situation  is,  of 
course,  to  let  X  represent  the  weight  of  the  bath,  and  work  out 
the  amount  of  iron  reduced  into  the  bath  by  the  two  methods, 
getting  the  result  expressed  in  each  case  in  terms  of  X.  Since 
the  two  expressions  represent  the  same  quantity,  we  put  them 
equal  to  each  other  and  solve  for  X.  Having  obtained  X,  we 
substitute  it  in  either  expression  and  get  the  result  asked  for. 

Let  X  be  the  weight  of  the  bath  after  the  reaction,  in  pounds; 
the  bath  before  the  reaction  weighs  100,000  pounds. 

Oxidized  out : 

Carbon  (100,000X0.035)— X  0.03  =  3,500—0.03  X  pounds. 

Silicon  100,000X0.02  =  2,000 

Phosphorus  100,000X0.0075  =      750 

Manganese  100,000X0.005  =      500 

Oxygen  required: 

For  carbon  (3,500—0.03  X)  16/12  =  4,677—0.04  X  pounds. 

For  silicon  2,000X32/28  =  2,286 

For  phosphorus  750X80/62  =      968 

For  manganese  500X16/55  =      145 


Sum  =  8,076 — 0.04  X  pounds. 
Oxygen  supplied: 

If  all  Fe2O3  of  ore  (27,000  pounds)  were  reduced  to 
FeO  =  27,000X16/160  =  2,700  pounds. 

If  all  Fe2O3  of  ore  were  reduced  to  Fe  27,000X48/160  = 
8,100  pounds. 

We  thus  see  that  we  will  certainly  require  more  oxygen  than 
the  Fe2O3  can  give  up  in  becoming  all  FeO,  but  not  as  much 
as  would  be  given  up  if  it  all  became  Fe.  If  all  the  Fe2O3  were 
considered  first  reduced  by  the  reaction  to  FeO,  giving  up 
2,700  pounds  of  oxygen,  the  reaction  will  still  require  the  fur- 
nishing of 

(8,076—0.04  X)— 2,700  =  5,376—0,04  X  pounds. 
of  oxygen,  which  would  have  to  be  furnished  by  FeO  becoming 


EFFICIENCY  OF  OPEN-HEARTH  FURNACES.          427 

Fe.  In  that  reduction,  however,  72  parts  of  FeO  gives  up  16 
of  oxygen  in  becoming  56  Fe.  The  reduced  Fe  will  be,  there- 
fore, 56/16  of  the  weight  of  oxygen  thus  furnished,  and  therefore, 

Reduced  Fe  =  56/16  (5,376—0.04  X)  pounds. 
=  18,816— 0.14  X  pounds. 

The  same  quantity  is  obtained  more  directly  as  follows: 
Fe  in  original  bath  100,000X0.9325       =  93,250  pounds. 
Fe  in  bath  after  reaction  =       0.97  X  pounds. 

Therefore,  Fe  reduced  =      0.97  X— 93,250  Ibs. 

These  two  expressions  represent  the  same  thing,  and,  therefore, 

18,816— 0.14  X  =  0.97  X— 93,250 
Whence  X  =  100,960 

And  the  reduced  iron  =  0.97  X— 93,250  =  4,681  pounds.     (1) 
(2)  The  ore  used  contains  altogether 

Fe  =  27,000  X 112/160  =  18,900  Ibs. 

Fe  reduced  to  metallic  state  =     4,681    " 


Fe  remaining  in  slag  as  FeO  =  14,219  " 

Weight  of  FeO  =  14,219x72/56  =  18,282  "  =  61.1  per  cent. 

Weight  of  P2O5  =  750  +  968  =     1,718  "  =    5.7       " 

Weight  of  MnO=  500+145  =       645  "  =    2.1 

Weight  of  CaO  =    2,000  "  =    6.7 

Weight  of  SiO2  =3,000  +  4,286      =     7,286  "  =24.4 

Total  weight  of  slag  =  29,931    "  (2) 

(3)  The  physical  data  available  do  not  permit  of  calculating 
the  actual  heat  of  the  reaction  at  1,300°,  since  many  of  the 
specific  heats  needed  are  lacking.  We  will  therefore  foot  up 
the  items  of  heat  evolution  and  absorption  unconnected  for 
temperature,  which  is  the  only  thing  to  be  done  under  the  cir- 
cumstances. 

Heat  Evolution.  Calories 

Si  to  SiO2  2,000X7,000  =14,000,000 

PtoP2O5  750X5,892  =    4,419,000 

MntoMnO  500X1,653  =        826,500 

CtoCO  471X2,430  =     1,144,500 

SiO2  to  FeO.  SiO2  7,286  X     144  =     1,049,200 

CaO  to  3CaO.  P2O5  2,000  X    949  =     1,898,000 


Total  =  23,337,200 


428  METALLURGICAL  CALCULATIONS. 

Heat  Absorption. 

Calories. 

Fe2O3toFeO  18,900  X    573          =10,829,700 

FeOtoFe  4,681X1,173          =    5,490,800 

Fe3CtoFe3  +  C  471 X    705  (?)    =       332,000  (?) 

FeSi  to  Fe  +  Si  2,OOOX    931  (??)  =    1,682,000  (??) 

Fe3P  to  Fe3  +  P  750X1,400  (??)  =     1,050,000  (??) 


Sum  =  19,564,500 

These  calculations,  therefore,  show  a  minimum  surplus  of 
heat  in  the  reaction  of  23,337,200  -  19,564,500  =  3,772,700  Cal., 
an  amount  which  would  raise  the  temperature  of  the  slag  and 
resulting  metal  approximately  100°  C.  above  the  1,300°  at  which 
the  reacting  materials  came  together.  The  quantity  above 
marked  (?)  is  a  little  doubtful  in  amount,  but  those  marked 
(??)  are  very  doubtful,  perhaps  may  not  exist  at  all.  If  these 
are  omitted  from  the  heat '  absorption  the  surplus  heat  is  in- 
creased some  50  per  cent,  of  its  value,  and  the  rise  in  tem- 
perature might  be  in  the  neighborhood  of  150°.  Further,  some 
of  the  CO  formed  by  the  oxidation  of  carbon  may  be  burned  to 
CO2  close  to  the  surface  of  the  bath,  by  free  oxygen  in  the  furnace, 
still  further  increasing  the  rise  in  temperature. 

The  conclusion  from  these  calculations  and  discussions  is 
that  the  pig  iron  and  ore  reaction  in  the  Monell  process  is  a 
heat  evolving  reaction,  which,  independently  of  the  heating 
effect  of  the  fuel  used  by  the  furnace,  could  increase  the  tem- 
perature of  the  contents  of  the  furnace  at  least  100°,  and  pos- 
sibly 150°.  It  would  be  highly  interesting  to  have  a  typical 
heat  such  as  this  followed  closely  with  a  good  reliable  pyro- 
meter, so  as  to  check  the  indications  of  the  thermochemical 
study  of  the  process. 


CHAPTER  XII. 
THE  ELECTROMETALLURGY  OF  IRON  AND  STEEL. 

Electrical  methods  may  enter  into  the  extraction  of  a  metal 
from  its  ores  either  as  electrolytic  or  as  electrothermal  pro- 
cesses. Electrolytic  processes  are  those  in  which  the  electric 
current  is  used  for  its  electrolytic  action,  i.e.,  for  its  electrical 
decomposing  and  depositing  properties;  electrothermal  pro- 
cesses are  such  as  use  the  current  merely  as  a  source  of  heat, 
to  furnish  the  sensible  heat  and  high  temperature  necessary 
for  melting  materials  or  for  bringing  about  chemical  reactions. 
So  far  electrolytic  processes  have  entered  the  metallurgy  of 
iron  only  as  used  by  Burgess  for  electrolytically  refining  nearly 
pure  iron  in  an  aqueous  electrolyte  and  depositing  chemically 
pure  iron;  the  electrolysis  of  fused  iron  salts  has  not  been  prac- 
tically utilized.  Up  to  the  present,  electrothermal  processes 
are  in  commercial  use  for  melting  together  wrought  iron  and 
cast  iron  to  make  steel,  also  for  keeping  cast  iron  melted  while 
its  impurities  are  being  extracted  by  oxidation ;  the  electro- 
thermal reduction  of  iron  ores  to  cast  iron  has  been  proved 
technically  possible,  and  may  in  some  places  prove  commer- 
cially practicable. 

ELECTROTHERMAL  REDUCTION  OF  IRON  ORES. 
If  the  electric  current  is  used  to  furnish  the  heat  energy 
necessary  to  reduce  iron  ore,  it  cannot  displace  the  reducing 
agent — carbon.  In  ordinary  blast  furnace  practice  the  carbon 
is  first  burned  to  provide  the  heat  necessary  to  smelt  down  the 
pig  iron  and  slag,  and  the  product  of  this  incomplete  combus- 
tion— CO — abstracts  oxygen  from  the  ore.  The  two  equations 
are  practically: 

At  the  tuyeres — 

30C  +  15O2  +  57.15  N2  =  30CO  +  57.15  N2 

Reduction : 
4Fe203  +  30CO  +  57.15  N2  =  12CO2+18CO-r57.15  N2  +  8Fe 

429 


430  METALLURGICAL  CALCULATIONS. 

These  equations  show  us  that  to  produce  8Fe  =  448  parts, 
30C  =  360  parts  of  carbon  is  the  minimum  necessary,  which  is 
first  burned  to  CO  at  the  tuyures,  and  then  the  producer  gas 
thus  formed  (N2  and  CO)  reduces  the  iron  oxides  above.  If 
the  heat  for  fusion  is  furnished  electrically,  the  first  combus- 
tion is  unnecessary,  all  blowing  in  of  air  is  dispensed  with  and 
the  reaction  taking  place  is: 

14Fe203  +  30C  =  12C02+18CO  +  28Fe 

And  we  have  28Fe  =  1,568  parts  reduced  by  30C  =  360  parts, 
a  consumption  of  less  than  one-third  as  much  carbon  as  is 
required  in  the  blast  furnace.  The  operation  consists,  therefore, 
in  mixing  iron  ore  and  carbon  so  that  for  every  part  of  iron 
present  about  0.25  parts  of  carbon  is  present,  using  the  proper 
quantity  of  limestone  or  other  material  to  flux  the  gangue  of 
the  ore  to  a  fusible  slag,  and  then  furnishing  electrically  the 
heat  necessary  to  cause  the  chemical  reaction,  melt  down  the 
resulting  iron  and  slag,  and  supply  radiation  losses.  The 
gases  resulting  from  this  electrical  reduction  are  combustible, 
just  as  the  gases  from  the  blast  furnace,  and  since  there  is  no 
blast  to  be  heated  they  can  very  well  be  utilized  to  preheat  the 
charges  coming  into  the  furnace,  and  thus  save  some  of  the 
electrical  energy  needed. 

Problem  76. 

A  magnetite  ore  contains: 

Per  Cent.  Per  Cent. 

Fe2O3 60.74        MgO. 5.50 

FeO 17. 18        P205 . .0.04 

SiO2 6.60        S 0.57 

A12O3 1.48        CO2 2.05 

CaO 2.84         H2O 3.00 

It  is  to  be  mixed  with  pure  carbon  (charcoal  fines)  and  a 
suitable  flux;  the  fixed  carbon  being  0.25  per  cent,  of  the  iron 
present;  the  flux  silica  sand,  so  as  to  make  a  slag  with  33  per 
cent,  of  SiO2.  Neglect  the  ash  and  assume  10  per  cent,  of 
moisture  in  the  charcoal.  Assume  also: 

(a)  The  pig  iron  to  contain  4  per  cent.  C,  3.5  per  cent.  Si, 
92.4  per  cent.  Fe. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          431 

(6)  The  slag  and  pig  iron  to  contain    at    tapping  600  and 
400  Calories  of  heat  respectively. 

(c)  The  heat  losses  by  radiation,  etc.,  to  be  30  per  cent,  of 
the  total  heat  requirement  of  the  furnace. 

(d)  The  hot  gases  to  escape  at  300°'  C. 

(e)  The  iron  to  be  completely  reduced  into  the  pig  iron. 
(/)  The  sulphur  to  go  entirely  into  the  slag  as  CaS. 

Requirements : 

(1)  The  weights  of  ore,  flux  and  charcoal  dust  needed  per 
1,000  kg.  of  pig  iron  produced. 

(2)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace. 

(3)  A  heat  balance  sheet  of  the  furnace. 

(4)  The  number  of  kilowatt   days   of  electrical  energy  re- 
quired per  metric  ton  of  pig  iron  produced. 

Solution : 

(1)  The  ore  must  supply  924  kg.  of  iron.     But  100  parts  of 
ore  contains  iron  as  follows : 

Kg. 

In  Fe203     60.74X112/160  =  42.52 
In  FeO       17.18X   56/72    =  13.36 


Sum  =  55.88  (1) 

Ore  required  per  1,000  kg.  of  pig  iron: 
924-0.5588  =  1,654kg. 

The  slag-forming  ingredients  from  this  amount  of  ore  are 
as  follows: 

Kg. 

APO31,654X  0.0148  =    24.6 

MgOl,654X  0.0550  =    91.0 
CaO  1,654  X  (0.0284— 0.0057X56/32) 

=  1,654X0.0184  =    30.4 

Si02(l,654x  0.0660)— (35x60/28)  =  109.2—75  =    34.2 

CaS  1,654  X  (0.0057X72/32)  =    21.2 


Sum  =  201.4 
If  x  parts  of  SiO2  sand  are  added  to  these,  the  total  weight 


432  METALLURGICAL  CALCULATIONS. 

of  slag  will  be  201.4+*,  of  SiO2  in  it  34.2+*.      And  since  the 
SiO2  is  to  be  33  per  cent,  of  the  weight  of  slag,  then 

34.2  +  *  =  0.33  (201.4  +  *) 
Whence  *  =     48  kg.  (1) 

The  charcoal  dust  used  must  contain  fixed  carbon  equal  to 
0.25  of  the  iron  present,  i.  e., 

924X0.25  =  231kg. 

And  since  it  is  90  per  cent,  fixed  carbon  the  dust  required  is: 
231-^-0.90  =  257kg. 


Charges. 
Ore  1654  Ki 

?• 
6 
2 
2 
6 

Pig  Iron. 
Fe     703 

2 

Slag. 

Gases. 
o           am 

4 
2 
0 

Fe203 
FeO.. 

1004. 

284. 

Fe...... 

Si 

.221 
35 

.0 
.0 

SiO, 

34  2 

O  
O 

..63. 
.  40 

SiO2 

...109. 

A1203 
CaO.. 
MgO.. 
P205.. 
S  
CO, 

.  .  .24. 

ALO, 

24  6 

46. 
91. 
0. 
9. 
33. 

9 
0 
6 
4 
Q 

p  

...0 

.3 

i 

CaO   . 

.30.4 

O 

.  .  .4. 

7 
3 

9 
6 

MgO.... 
Ca  
S  

..91.0 
..11.8 
...9.4 

0..... 

CO,'.'.'. 

H,0.. 

...0. 

..33. 
...49. 

•WVJ   •    • 

H20.. 

..49. 

6 

Flux 48  Kg. 

SiO2 48.0  SiO2 48.0     

Charcoal 257  Kg. 

C 231.0  C 40.0     C 191.0 

H2O 26.0  H2O 26.0 


1959.0  999.5  249.4  118.1 

(3)  Heat  available  is  the  heat  of  oxidation  of  carbon  plus 
that  furnished  by  the  electric  current.  The  charge  gives  up 
to  the  carbon,  as  shown  on  the  balance  sheet,  301.4  +  63.2  + 
40.0  +  4.7  +  0.3  =  409.6  kg.  of  oxygen.  The  191  kg.  of  carbon 
burned  would  take  191X16/12  =  254.7  kg.  of  oxygen  to  burn 
it  to  CO,  leaving  154.9  kg.  of  oxygen  to  burn  CO  to  CO2.  This 
would  burn  154.9x28/16  =  271.1  kg.  of  CO  to  CO2. 

The  heat  of  formation  of  the  slag  may  be  taken  as  approxi- 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          433 

mately  150  Calories  per  kg.  of  contained  Si02  +  AP03.  The 
heat  of  combination  of  carbon  with  the  iron  is  a  doubtful  quan- 
tity, which  may  be  taken  at  705  Calories  per  kilogram  of  car- 
bon. The  formation  of  CaS  gives  2,947  Calories  per  kilogram 
of  sulphur. 

Letting  the  heat  furnished  by  the  electric  current  =  x, 
and  neglecting  the  heat  of  oxidation  of  the  small  amount  of 
electrode  carbon  consumed,  we  have: 

Heat  Available. 

Calories. 

Supplied  by  electric  current :  % 

Oxidation  of  C  to  CO  191 X  2,430  =  434,130 

Oxidation  of  CO  to  CO2  271.1X2,430  =  658,770 

Formation  of  silicate  slag  106.8 X     150  =  16,020 

Formation  of  CaS  9.4x2,947  =  27,700 

Formation  of  Fe3C  40.   X    705  =  28,200 


Sum  =  1,194,820-f* 

Heat  Distribution. 

Calories. 
Reduction  of  Fe  from  Fe2O3 

703.2X1,746  =  1,229,790 

Reduction  of  Fe  from  FeO 

221.0X1,173  =      259,230 

Reduction  of  Si  from  SiO2 

35.0X7,000  =     245,000 

Reduction  of  P  from  P2O5 

0.3X5,892  =         1,770 

Reduction  of  Ca  from  CaO 

11.8X3,288  =       38,800 

Expulsion  of  CO2  from  ore 

33.9X1,026  =       34,780 

Evaporation  of  H2O  from  charges 

75.6X606.5  =       45,850 

Sensible  heat  in  gases,  at  300° 

CO  174.6  kg.  =  138  m3 

X0.311  =    42.9 


434  METALLURGICAL  CALCULATIONS. 

CO2  459.9  kg.  =  232  m3 

X  0.436  =  101.1 

H2O  75.6  kg.  =  93  m3 

X  0.385  =    35.8 


179.8X300  =       53,940 

Sensible  heat  in  slag  249.4x600  =  149,640 

Sensible  heat  in  pig  iron  1,000X400  =  400,000 

Loss  by  radiation,  etc.,  0.30  (1,194,820  +  *)  =  358,450 +  0.3* 

Sum  total  =  2,817,250  +  0.3* 
Equating  the  heat  available  and  accounted  for  we  have: 

1,194,820  +  *  =  2,817,250  +  0.3  x. 
Whence  x  =  2,317,760  Calories. 

And  the  sum  total  of  heat  requirement  is 

3,512,580  Calories, 
of  which  the  electric  current  supplies 


=  0.66  =  66  per  cent.  (3) 

(4)  A  kilowatt-day  of  electricity  is  equal  to 

0.239X60X60X24  =  20,650  Calories. 
There  is,  therefore,  required  per  ton  of  pig  iron  produced: 

2,317,760 


20,650 


=  112  kilowatt  days.  (4) 


This  figure  might  be  materially  reduced  by  using  the  waste 
gases  to  warm  up  the  charges  entering  the  furnace.  It  is  also 
possible  that  in  a  properly  designed  shaft  the  gases  passing  out 
might  contain  nearly  equal  volumes  of  CO  and  CO2,  instead  of 
3CO  to  2CO,  as  assumed  in  this  problem,  from  best  ordinary 
blast  furnace  practice.  Any  greater  utilization  of  the  heat- 
producing  power  of  the  carbon  would  decrease  the  electrical 
energy  required;  the  above  calculated  value  is  given  as  a  safe 
figure  for  this  ore  on  which  to  base  working  calculations. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL. 
Problem  77. 


435 


At  Sault  Sainte  Marie,  Canada,  roasted  pyrrhotite  ore  was 
smelted  with  the  addition  of  limestone  and  charcoal  fines.  The 
mixture  used  contained  400  pounds  of  ore,  110  pounds  charcoal 
dust  and  85  pounds  limestone.  The  analyses  of  each  of  these 
materials  was: 


Roasted  Ore. 

Fe2O3 65.43 

CuO .   0.51 

NiO 2.84 

SiO2 10.96 

A12O3..  .   3.31 


CaO, 


3.92 


MgO ,  ...   3.53 

SO3 * 

P205 

H2O.. 


3.90 
0.03 
5.57 


Limestone. 

CaO 52.00 

MgO 2.10 

Fe2O3 0.60 

A12O3 0.21 

SiO2 1.71 

P2O5 0.01 

SO3 0.13 

CO2..  ..43.15 


Charcoal  Dust. 

Fixed  C 55.90 

Vol.  matter 28.08 

Moisture 13.48 

Ash..  .   2.54 


Using  165.65  kilowatts  effective  electric  energy  for  56  hours 
20  minutes,  there  was  produced  7,336  pounds  of  nickeliferous 
pig  iron  and  5,062  pounds  of  slag,  having  the  following  average 
compositions: 


Slag. 

SiO2 16.44 

A12O3 13.86 

CaO 42.87 

MgO 8.80 

CaS 13.34 

FeO 0.84 

Undetermined . .          .3.85 


c  

Pig  Iron. 
3.05 

Si       

5.24 

s  

.  0.01 

P       

0  .  05 

Cu       

0.81 

Ni.. 

.   3.94 

Fe..  86.90 


Requirements : 

(1)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace. 

(2)  A  heat  balance  sheet  of  the  furnace,  making  necessary 
assumptions  where  data  is.not  furnished. 

(3)  The  thermal  efficiency  of  the  furnace. 
Solution : 

(1)  The  amount  of  ore  used  may  be  calculated  either  from 
the  iron,  the  nickel  or  the  copper.  The  nickel  and  copper  are 
the  easiest  to  use,  because  there  is  supposed  to  be  none  of 


436  METALLURGICAL  CALCULATIONS. 

them  in  the  slag,  but  they  are  the  least  reliable,  because  present 
in  such  small  amount.  The  pig-metal  contains  7,336X0.81  = 
59.4  pounds  of  copper,  and  since  the  roasted  ore  contains  0.51 

f»ty   /» 

X  7Q'     =  0.41   per  cent,   copper,   the   weight  of  ore  used,  on 

this  basis,  would  be  59.4^-0.0041  =  14,493  pounds.  As  for 
nickel,  the  pig-metal  contains  7,336X0.0394  =  289  pounds  of 

nickel,   and  since  the  roasted  ore  contained   2.84X^  =  2.23 

75 

per  cent.,  the  weight  of  this  used  should  have  been  289  -i-  0.0223  = 
12,960  pounds.  These  figures  differ  so  much  that  we  will 
make  the  calculation  on  the  basis  of  the  iron.  There  is  iron 
present  as  follows : 

Pounds. 

In  the  pig  iron  7,336X0.8690  =  6,375 

In  the  slag  5.062  X  0.0084  X  56/72  =       33 


Total  in  products  =  6,408 


In  100  pounds  of  ore  65.43X0.7  =  45.80 

In  21  pounds  limestone  21.  X  0.0060X0.7    =       .08 
In  27.5  pounds  of  charcoal  27.50  X  0.0025 

X0.7  =       .04 


In  charge,  per  100  pounds  of  ore  used  =  45.92 

Therefore,  ore  necessary  to  supply  the  iron  in  products: 

Pounds. 

6,408-^0.459  =  13,961 

with  which  will  be  used : 

Charcoal     =  13,961x110/400  =     3,839 

Limestone  =  13,961X85/400  =    2,967 

We  are  now  ready  to  construct  the  balance  sheet  as  soon 
as  we  assume  probable  values  for  the  composition  of  the  vola- 
tile matter  and  ash  of  the  charcoal.  The  ash  might  be  taken 
as  containing  on  an  average:  K2O  15  per  cent.,  CaO  40,  MgO 
20,  MnO  15,  and  Fe203  10  per  cent.  The  volatile  matter  is  due 
to  insufficient  charring,  and  the  gases  given  off  on  heating 
may  be  assumed  as,  by  volume,  CO2  25  per  cent.,  CO  15,  H2 


ELECTROMETALLURGY  OF  IRON  AND  STEEL. 


437 


50,  CH4  10.  This  would  make  the  volatile  matter  to  contain,  in 
per  cents,  by  weight,  carbon  33.7,  oxygen  58.5,  hydrogen  7.  8 
per  cent.  (The  verification  of  this  last  statement  is  a  nice 
little  exercise  in  chemical  arithmetic.)  The  full  statement  of 
the  elementary  composition  of  the  charcoal,  for  use  in  making 
the  balance  sheet,  is  therefore: 

Per  Cent. 

Fixed  carbon 55.90 

Volatile  carbon 9.46  ] 

Volatile  hydrogen 2.19  I  28.08% 

Volatile  oxygen 16.43  J 

Moisture 13.48 

K20 0.38 

CaO 1.02 

MgO 0.51  }    2.54% 

MnO 0.38 

Fe203 0.25 

Balance  Sheet,  per  7,336  Lbs.  Pig-metal. 

Charges.  Pig-metal.  Slag.  Gases. 

Roasted  Ore.  .13,961 

Fe2O3 9,135     Fe 6,375  FeO. 25     O 2,375 

NiO 397     Ni 289  NiO 28     0 80 

CuO 71     Cu 59  0 12 

SiO2 1,530     Si 384  SiO2 1,091     O 55 

A12O3 461      A12O3 461      

CaO 547      CaO 166     0 109 

MgO 493      .- MgO 493      

SO3 545     S 1  CaS 490     0 327 

P2O5 4     P 4 0 0 

H20 778  H20 778 

Limestone.  . .  2,967 

CaO 1,543      CaO 1,542  0 0 

MgO 62      MgO 62  

Fe2O3 18     FeO 16  0 2 

A12O3 6      A1203 6  

SiO2 51      SiO2... 51  

P2O5 0      P2O5 0  

SO3 4      CaS 4  

CO2 1,280  CO.     ..1280 


Charcoal  dust  3,839 


438  METALLURGICAL  CALCULATIONS. 

• 

Fixed  C 2,146     C 224      C 1,922 

Vol.  C 363 

Vol.  H 84 

Vol.  O 631 

H20 518 

K20 15 

CaO 39 

MgO 20 

MnO 15 

Fe2O3..  10 


C  

.  .  .363 

H  

.    .  .84 

O 

.  .  .631 

H  O 

518 

K2O 

15 

.    .         CaO 

39 

MgO  . 

20 

MnO 

.  .  .15 

FeO.. 

.  9 

O.. 

.    1 

Electrode..  66 


66      C..  ..66 


20,828  7,336  4,533  8,962 

There  is  a  lack  of  close .  correspondence  between  the  weights 
and  compositions  of  slag,  as  observed  and  as  calculated,  due 
evidently  to  inaccurate  sampling  and  analyses  of  the  roasted 
ore  and  slag. 

(2)  From  the  balance  sheet  we  can  deduce  the  heat  evolved 
and  absorbed  in  the  chemical  reactions  in  the  furnace.  The 
more  involved  items  are  calculated  as  follows: 

Oxidation  of  carbon  to  CO :  All  the  carbon  put  into  the  fur- 
nace as  fixed  carbon  goes  out  as  either  CO  or  CO2,  except  that 
going  into  the  pig  iron.  The  carbon  burnt  to  CO  in  the  fur- 
nace may  be,  therefore,  taken  as  1,922  +  66  =  1,988  pounds, 
evolving  1,988X2,430  =  4,830,800  pound-Calories. 

Oxidation  of  CO  to  CO2:  There  is  given  up  in  the  furnace, 
by  the  reductions  accomplished,  3,021  pounds  of  oxygen,  of 
which  1,988x16/12  =  2,651  pounds  would  burn  fixed  carbon 
to  CO,  as  above  shown,  leaving  370  pounds  to  burn  CO  to  CO2. 
This  would  oxidize  370X28/16  =  648  pounds  of  CO  to  CO2, 
which  would  evolve  648X2,430  =  1,574,300  pound-Calories. 

Heat  energy  of  electric  current:  One  kilowatt-second  is  0.239 
kilogram  Calories,  or  0.527  pound-Calories.  The  current  being 
on  56  hours,  20  minutes,  or  202,800  seconds,  the  heat  equiva- 
lent of  the  current  used  is: 

0.527X202,800X165.65  =  17,704,000  pound-Calories. 

Heat  in  escaping  gases:  We  are  here  confronted  with  the 
fact  that  no  observation  of  the  temperature  of  these  was  given. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          439 

There  is  no  essential  reason  why  they  should  escape  very  hot 
from  the  furnace,  if  properly  run  and  conducted,  so  we  will 
assume  a  maximum  temperature  of  500°  C.  The  gases  would 
consist  of  0,958  +  2,651-648  =  4,061  pounds  of  CO  and  648  + 
370  =  1,018  pounds  of  CO2,  from  the  oxidation  of  fixed  carbon 
in  the  furnace;  plus  1,280  of  CO2  from  the  limestone,  and  666 
pounds  CO2,  254  pounds  CO,  97  pounds  CH4  and  60  pounds 
of  H2  from  the  volatile  matter  of  the  charcoal.  To  these  must 
be  added  1,290  pounds  of  water  vapor.  The  heat  is  therefore 

CO  4,315  Ibs.  =  54,800  ft3 X  0.304=  16,650  oz.  Cal. 

CO2  2,964  Ibs.  =  23,9hu  lt°X0.480  =  11,500 
CH4  97  Ibs.  =    2, 150  ft3  X  0.490=    1,050       " 

H2  60  Ibs.  =  10,700  ft3 X  0.304=   3,250       " 

H2O  1,290  Ibs.  =  25,500  ft3 X 0.415=  10,550       " 


Sum  =117,100  ft3  43,000  per  1° 

=  21,500,000       "         per  500 
=    1,343,750  Ib.-Cal. 

The  other  items  of  the  heat  balance  sheet  are  almost  self- 
explanatory,  and  the  complete  balance  is  as  follows: 

Heat  Available. 

Pound-Col. 

Energy  of  the  electric  current  =  17,704,000 

Oxidation  of  C  to  CO  =  4,830,800 

Oxidation  of  CO  to  CO2  =  1,574,600 

Combination  of  C  with  Fe3  224x705  157,900 

Combination  of  Ca  with  S  220X2,947  =  .648,300 
Formation  of  silico-aluminate  slag 

(SiO2  +  APO3)  1609X150  =  241,400 

Total  =  25,157,000 
Heat  Distribution. 
Reductions : 

•Fe2O3  to  Fe  6,375X1,746  =  11,130,750 

NiO  to  Ni  289X1,051  =        303,750 

CuO  to  Cu  59 X    593  =          35,000 

SiO2  to  Si  384  X  7,000  =    2,688,000 

SO3  to  S  218X2,872  =       626,100 

P2O5  to  P  4X5,892  =         23,550 

CaO  to  Ca  274x3,288  =        900,900 

Fe2O3toFeO  50  X    446=         22,300 


440  METALLURGICAL  CALCULATIONS. 

Expulsion  of  CO2  from  flux: 

1,280X1,026  =  1,313,300 

Evaporation  of  H2O                   1,290X606.5  =  783,400 

Sensible  heat  in  gases                                          =  1,343,750 

Sensible  heat  in  pig  iron            7,336  X     400  =  2,934,400 

Heat  in  slag                                 4,533  X    600=  2,719,800 

Loss  by  radiation,  conduction,  etc.                  =  332,000 


Total  =  25,157,000    (2) 

(3)  The  essential  work  done  by  the  furnace  is  the  reduc- 
tions, evaporation  of  moisture  and  decomposition  of  carbon- 
ates. The  heat  in  slag,  iron,  gases  and  radiation  loss  are  all 
susceptible  of  diminution  or  of  being  more  or  less  returnable 
to  the  furnace.  The  usefully  applied  heat  is,  therefore,  17,- 
827,050  pound-Calories.  To  produce  this  there  was  consumed 
the  25,157,000  pound-Calories  actually  generated,  and  there 
was  wasted  13,390,000  pound-Calories,  the  calorific  power  of 
the  gases  escaping  from  the  furnace,  which  should  have  been 
generated  or  might  be  utilized,  making  a  total  of  38,527,000 
pound-Calories  disposable. 

The  working  thermal  efficiency  over  all  was  therefore: 

17,827,050 

38,527.000  =  °'46  =  46  P 

PRODUCTION  OF  STEEL. 

There  are  three  methods  of  producing  steel  electrically  which 
are  practicable.  First,  the  electric  furnace  may  replace  the 
crucible  simply  as  a  melting  apparatus,  in  producing  a  cast 
steel  from  cemented  bars;  second,  the  electric  furnace  may  re- 
place the  crucible  or  open-hearth  furnace  as  an  apparatus  in 
which  to  melt  together  wrought  iron  and  pure  cast  iron,  such 
as  washed  pig  metal,  although  in  this  operation,  the  electric 
furnace  is  more  like  the  crucible  method,  in  that  there  is  not 
necessarily  any  oxidation  of  the  metal  or  of  its  impurities  in 
the  operation;  third,  the  electric  furnace  may  be  used  to  melt 
or  keep  melted  cast  iron,  while  its  impurities  are  oxidized  out 
by  additions  of  iron  ore,  in  this  operation  resembling  the  Ucha- 
tius  method  of  making  crucible  steel  or  the  "  pig  and  ore  " 
process  of  making  steel  in  the  open-hearth  furnace. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          441 

The  particular  advantages  possessed  by  the  electric  furnace 
processes  are,  compared  with  the  crucible  process,  the  larger 
quantities  in  which  the  steel  can  be  made  in  one  operation,  the 
absence  of  carbon  in  the  furnace  lining,  thus  controlling  better 
the  carbon  and  silicon  in  the  steel,  and  the  higher  temperature 
enabling  a  more  basic  slag  to  be  kept  fluid  and  the  sulphur  to 
be  better  eliminated;  the  advantages  compared  with  the  open- 
hearth  furnace  are  the  absence  of  gases  of  combustion  in  con- 
tact with  the  metal,  and  the  higher  temperature  available, 
which  permits  of  very  basic,  refractory  slags  being  made  and 
kept  thinly  fluid,  and  thus  gives  better  control  of  sulphur  and 
phosphorus.  In  addition  to  these,  in  both  cases,  may  be 
mentioned  the  commercial  advantages,  for  the  saving  in  cru- 
cibles alone  makes  the  electric  furnace  superior  in  this  respect 
to  the  crucible  process,  and  the  electric  furnace  can  compete 
successfully  as  regards  cost  with  the  regenerative  open-hearth 
furnace  wherever  water  power  costs  less  than  $10.00  per  horse- 
power-year and  coal  costs  over  $5.00  per  ton. 

A  particular  point  to  be  noted  is,  that  when  heating  by  com- 
bustion is  used  the  efficiency  of  the  absorption  of  heat  by  the 
charges  decreases  very  rapidly  as  the  temperature  gets  higher. 
For  instance,  if  a  cold  ingot  of  iron  is  placed  in  a  furnace  the 
temperature  of  which  is  1,500°,  the  iron  absorbs  heat  with  very 
great  rapidity  from  the  start  up  to,  say,  1,000°;  but  with  de- 
creasing rapidity  thereafter.  The  rate  of  transfer  of  heat 
from  the  gases  to  the  iron  is  proportional  to  the  difference  of  tem- 
perature, and  is  some  fifteen  times  as  fast  when  the  iron  is 
at  0°  as  when  it  is  at  1,400°.  If  the  efficiency  of  the  heating 
by  furnace  gases  is,  say,  25  per  cent,  in  bringing  metal  up  to 
1,500°,  it  is  likely  that  the  distribution  of  this  efficiency  is  pro- 
portioned about  as  follows: 

Heating  from  0°  to  500° .  45  per  cent,  efficiency. 

Heating  from  500°  to  1,000° 27  per  cent,  efficiency. 

Heating  from  1,000°  to  1,500° 3  per  cent,  efficiency. 

On  the  other  hand,  the  conversion  of  electrical  energy  into 
heat  in  the  substance  of  the  material  to  be  heated  is  not  a 
contact  or  transfer  phenomenon,  but  a  thermodynamically 
frictionless  transfer  of  100  per  cent,  efficiency,  and  equally  so  at 
the  highest  as  at  the  lowest  temperatures.  Only  radiation  and 


442  METALLURGICAL  CALCULATIONS. 

conduction  losses  need  be  considered,  the  problem  is  not  how 
much  heat  can  you  get  into  a  body  but  how  much  can  you 
keep  in;  it  is  already  all  in,  100  per  cent,  of  it,  to  start  with. 
In  a  large,  properly  designed  electric  furnace  the  radiation  and 
conduction  losses  of  heat,  even  working  to  the  highest  tem- 
peratures, can  be  kept  at  15  to  25  per  cent,  of  the  total  heat 
generated,  giving  an  efficiency  of  75  to  85  per  cent. 

It  may  very  well  be,  that  there  are  places  where  the  relative 
prices  of  coal  and  electric  power  are  such  that  coal  is  the  cheaper 
for  heating  to  500°,  at  45  per  cent,  efficiency,  or  to  1,000°  at 
36  per  cent,  efficiency,  or  even  to  1,500°,  at  25  per  cent,  effi- 
ciency, but  that  the  combination  of  heating  by  fuel  to  1,000° 
at  36  per  cent,  efficiency  and  then  by  electricity  from  1,000°  to 
1,500°,  at  70  per  cent,  efficiency,  would  be  the  cheaper  plan, 
or  even  by  fuel  to  500°,  at  45  per  cent,  efficiency,  and  then 
by  electricity  from  500°  to  1,500°,  at,  say,  75  per  cent,  efficiency, 
would  be  commercially  advantageous. 

Illustration:  Steel  bars  are  to  be  melted  in  an  electrical 
furnace.  It  takes  300  Calories  effective  in  the  steel  per  kilo- 
gram to  heat  it  to  a  tapping  heat;  the  electric  furnace  supplies 
this  at  a  net  thermal  efficiency  of  75  per  cent.  To  heat  the  bars 
to  750°,  cherry  red,  without  melting  them,  requires,  88  Calories, 
or  29  per  cent,  of  the  total.  If  the  bars  were  heated  in  a  coal 
furnace  to  750°,  and  then  transferred  to  the  electric  furnace, 
some  25  per.  cent,  of  the  electrical  power  might  be  saved.  If 
this  heating  were  done  by  coal  having  a  calorific  power  of 
8,500,  at  a  thermal  efficiency  of  25  per  cent.,  there  would  be 
needed  40  grams  of  coal.  The  question  is,  therefore,  the  rela- 
tive cost  of  .88 -v- 0.75  =117  Calories  delivered  electrically,  and 
40  grams  of  coal.  The  former  requires  117-^0.239  =  490 
kilowatt-seconds  =  8.2  kilowatt-minutes  =  0.14  kilowatt-hours. 
At  $10.00  per  kilowatt-year  (8,760  hours)  this  would  cost  0.14 
X  0.1 14  =  0.016  cents.  At  $5.00  per  metric  ton  the  coal  would 
cost  0.040X0.5  =  0.020  cent.  Under  such  assumed  conditions 
of  cost  of  power  and  of  coal  the  electrical  heating,  even  up 
to  750°,  would  be  the  cheaper. 

Problem  78. 

In  an  induction  electric  furnace  of  170  kilowatts  capacity, 
4.7  tons  of  steel  is  made  per  day  by  melting  together  cold- 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          443 

» 

washed  pig  iron  and  scrap  iron,  the  melted  steel  carrying  350 
Calories  per  kilogram. 
Required  : 

(1)  The  electric  energy  in  kilowatt  hours  required  per  ton 
of  steel  produced. 

(2)  The  thermal  efficiency  of  the  furnace. 

(3)  If  one-third  the  material  used  were  put  into  the  furnace 
melted,  carrying  275  Calories  per  kilogram,  what  would  be  the 
production  per  day  and  the  power  required  per  ton  of  steel? 

Solution  : 

(1)  Energy  for  4.7  tons  =          170  kw-days. 
Energy  for  1.0  ton  36.2 
Energy  for  1.0  ton  =         0.10  kw-year 
Energy  for  1.0  ton  =         869  kw-hours. 

(1) 

(2)  1  kw-hour  =  0.239X60X60     =  860       Calories 
869  kw-hours,                                                 747,340      " 
Heat  in  1  ton  of  steel  =  350  X  1,000=          350,000 


Thermal  efficiency  =          '        =  0.47  =  47  per  cent.  (2) 


(3)  Heat  in  melted  material  used  per  kilogram  of  steel  pro- 
duced =  275X1/3  =  92  Calories. 

Heat  to  be  supplied  by  the  current  350  —  92  =  258  Calories. 
Production  per  day  under  these  conditions: 


4.7  X         =  6.4  tons.  (3) 

Relative  times  for  the  heats  =  1  to  0.74. 

Energy  required  per  ton  =  170-^6.4  =  26.6  kw-days. 

=  0.07  kw-year. 
=  638  kw-hours.          (3) 

Problem  79. 

An  electric  steel  furnace  running  at  full  heat,  and  contain- 
ing about  2,500  kg.  of  steel,  loses  by  radiation,  etc.,  250,000 
Calories  per  hour;  2,500  kg.  of  melted  pig  iron  is  run  into  the 
hot  furnace,  carrying  250  Calories  per  kilogram,  and  it  is  treated 
with  500  kg.  of  iron  ore,  previously  heated  to  500°  C.,  and  50 


FeO  .... 

3 

96 

MgO   . 

0 

17 

c 

0  11 

SiO2 

5, 

50 

SiO,  .  . 

3 

14 

Si 

0  11 

MnO... 

0 

63 

0 

18 

Mn. 

0.15 

ALO,.. 

..   0. 

76 

A12O,. 

0. 

32 

S  

.  .  .   0.02 

CaO.. 

23 

0. 

006 

P.. 

.   0.01 

444  METALLURGICAL  CALCULATIONS. 

kg.  of  limestone  added  cold.  The  steel  produced  carries  400 
Calories  per  kilogram  and  the  slag  600  Calories.  The  opera- 
tion lasts  1  hour.  Assume  the  following  composition  of  ma- 
terials used  and  made: 

Pig  Iron.  Iron  Ore.  Limestone.  Steel. 

Fe 96.656     Fe2O3 85.93     CaO 53.74     Fe 99.60 

C. 2.700 

Si 0.600 

Mn 0.025 

S 0.007 

P 0.012 

MgO 0.97     S 0.001 

The  bath  was  treated  by  the  final  addition  of  10  kg.  of  cold 
ferro-manganese,  carrying  80  manganese,   16  iron  and  4  car- 
bon.    The  steel  obtained  weighed  2,630  kg. 
Required : 

(1)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace. 

(2)  The  weight  and  percentage  composition  of  the  slag. 

(3)  A  balance  sheet  of  the  heat  received  and  distributed. 

(4)  The  net  power  required  to  run  the  furnaces  and  the  cost 
of  power  per  ton  of  steel  made,  at  $25.00  per  kilowatt-year. 

Balance  Sheet. 

Charges.  Steel.  Slag.  Gases. 

Pig  Iron.... (2500  Kg.) 


Fe  

..2416.4 

2416.4 

C     

..      67.5 

2.5 

Si  

..     15.0 

2.9 

Mn..  
S  

..       0.6 
0.2 

0.5 

P.. 

0.3 

0.3 

C  

65  .  0 

SiO2  
MnO     

24.9     
0.8     

Ore (500  Kg.) 

Fe2O3 429.7  203.1  FeO 125.6  O 87.2 

FeO 19.8     FeO 19.8 

SiO2 27.5     SiO2 27.5  

MnO 3.2      MnO 3.2  

A12O3 3.8     A12O3 3.8  

CaO.., 11.2     CaO 11.2  

MgO 4.8     MgO 4.8  


ELECTROMETALLURGY  OF  IRON  AND  STEEL          445 


Limestone  

(50  Kg. 
26.9 
0  1 

}.  

CaO 

26. 

9     

CaO  

MeO 

MgO 

o 

1      

SiO2              

1.6 

SiO2  . 

...      1. 

6     

Fe2O3   

0.1 

FeO 

.  .     0. 

1      

A12O3 

0.2 

A12O3 

0. 

2     

CO2                 .    . 

21.2 

CO, 

21.2 

Ferro-manganese 
Fe 

(10  Kg 
1.6 
0.4 
8.0 

•) 
1.6 
0.4 
3.9 

C 

Mn 

MnO     .  . 

.  .  .     5 

3     

255. 

8 

3060.0 

2631.6 

173.4 

(2)  The  slag  contains : 

Per  Cent. 

SiO2 54.0  pounds  =  21.1 

A1203 4.0       "       =     1.5 

CaO 38.1       "       =  14.9 

MgO 4.9       "       =     1.9 

FeO 145.5       "       =  56.9 

MnO..  9.3       "       =    3.6 


255.8  =  99.9        (2) 

Heat  Available. 

Calories. 

Electric  current:  x 

Oxidation  of  C  to  CO  65.0X2,430  =  157,950 

Oxidation  of  CO  to  CO2  0.9X2,430  =      2,190 

Oxidation  of  Si  to  SiO2  12.1X7,000  =    84,700 

Oxidation  of  Mn  to  MnO  4.7  X  1,653  =      7,770 

Formation  of  slag  22.6  X    150  =      3,390 

Heat  in  melted  pig  iron  2,500  X    250  =  625,000 

Heat  in  iron  ore  500  X      77  =    38,500 

Sum  =  #+919,500 

Heat  Distribution. 

Calories 

Heat  in  melted  steel  2,630  X    400  =  1,052,000 

Heat  in  melted  slag  256  X     600  =      153,600 

Reduction  of  Fe2O3  to  FeO  386.7  X    446  =      172,470 


446  METALLURGICAL  CALCULATIONS. 

Reduction  of  FeO  to  Fe                           203.1  X  1,173  =  238,240 

Separation  of  carbon  from  iron                    65  X     705  =  45,800 
Heat  in  gas  at  1,500°: 

CO    =  151.7kg.  =  120.4  m3X 0.32X500  19,260 

CO2  =       1.4  kg.  =      0.7  m3X  0.60X500  200 

Loss  by  radiation,  etc.                                                       =  250,000 


Total  =  1,931,570 
Heat  to  be  supplied  by  current : 

x  =  1,931,570—919,500  =  1,012,070  Calories.  (3) 

(4)   One  kilowatt  furnishes  per  hour  860  Calories,  therefore 
the  power  required  to  run  the  furnace  is: 

1,012,070 -v- 860  =  1,177  kilowatts.  (4) 

At  $25.00  per  kilowatt-year  a  kilowatt-hour  would  cost 

$25.00-^-8,760  =  0.2854  cents. 
And  the  power  to  run  the  furnace  1  hour  would  cost 

0.002854X1,177  =  $3.36. 
And  the  cost  per  ton  of  steel: 

$3.36 -i- 2.630  =  $1.32. 


Problem  80. 

It  is  desired  to  design  a  plant  for  the  electro-deposition  of 
pure  iron  by  the  Burgess  process  (see  Transactions  American 
Electrochemical  Society,  Vol.  V.,  p.  201).  The  desiderata  and 
data  are  as  follows : 

Output,  25  metric  tons  per  day. 

Current  density,  110  amps,  per  square  meter. 

Anodes,  0.75X0.5  meters  immersed  X3m.m.  thick. 

Cathodes,  0.75X0.5  meters  immersed  Xl  m.m.  thick  at 
starting. 

Cathodes  to  be  run  until  deposit  is  1.5  c.m.  thick  on  each  side. 

Anodes  run  until  0.9  consumed. 

Tanks,  1.00  meter  deep,  0.6  m.  wide,  2  m.  long  inside,  filled  to 
within  0. 10  meter  of  top  with  electrolyte. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          447 

Working  distance  between  anode  and  cathode  6  centimeters 
at  starting. 

Electrolyte  contains  10  per  cent.  FeSO4.  7H2O,  and  5  per  cent. 
(NH4)2SO4,  specific  gravity  1.1,  electrical  resistivity  20  ohms 
per  centimeter  cube. 

Voltage  drop  in  connections  and  conducting  rods  0.3  volt  per 
tank. 

Main  conductors  carry  2  amps,  per  m.m.  square  of  section. 

Net  cost  of  electrical  power,  $25.00  per  kilowatt-year. 
Requirements : 

(1)  The  number  of  anodes  and  cathodes  per  tank  and  the 
number  of  tanks  in  the  plant  and  their  arrangement. 

(2)  The  weight  of  anodes  and  cathode  sheets,  increasing  the 
weight  of  immersed  part  10  per  cent.     Specific  gravity  of  the 
rolled  iron  7.9 

(3)  The  weight  of  ferrous  sulphate  and  ammonium  sulphate 
required  to  start  the  plant. 

(4)  The  drop  of  potential  across  the  electrodes  at  starting 
and  at  the  close  'of  a  deposition ;  the  drop  of  potential  per  tank ; 
the  total  voltage  needed  at  starting  the  plant  and  when  it  is 
in  regular  operation. 

(5)  The  cross  sectional  area  of  the  main  conductors. 

(6)  The  time  required  to  consume  an  anode  plate,  i.  e.t  in 
dissolving  iron  away  equal  to  0.9  of  its  weight. 

(7)  The  time  required  to  deposit  a  full  cathode  plate;  specific 
gravity  of  the  deposit  7.6 

(8)  The  electric  power  required  to  run  the  plant  and  its  cost 
per  ton  of  iron  deposited. 

Solution : 

(1)   110  amps,  per  square  meter  deposits  per  day: 

Cfi 

0.00001036X110X  yX60X60X24  =  2,757  grams  Fe. 

Therefore,  depositing  surface  required: 

25,000^-2,757  =  9,140  square  meters. 

Since  one  cathode  plate  has  a  depositing  area  on  both  sides 
of  0.75X0.5X2  =  0.75  square  meter,  the  number  of  cathode 
plates  required  in  the  whole  plant  is: 

9,140^0.75  =  12,187 


448  METALLURGICAL  CALCULATIONS. 

In  one  tank,  if  there  are  x  cathodes  and  x+l  anodes,  the 
thickness  of  these  plates  at  starting  is,  in  millimeters,  x  +  3 
(x+l)  =  4x  +  3  m.m.  The  number  of  spaces  between  anodes 
and  cathodes  being  2x,  and  each  of  these  being  6  c.m.  =  60 
m  m.  at  starting,  the  spaces  are  120#  m.m.  The  length  of 
the  tank  being,  inside,  2,000  m.m.  : 

124*  +  3  =  2,000 
whence  x  =  16.1. 

Each  tank  will  therefore  contain  16  cathodes  and  17  anodes, 
at  a  distance  apart,  at  starting  of 

2,000-16-3  (17) 


- 
=  6.04  c.m.  (1) 

Since  we  need  12,187  cathode  plates  we  need 
12,187-^16  =  761.  7  tanks 

Which  means  that  we  would  use  762  tanks.  (1) 

The  arrangement  of  the  tanks  in  series  and  groups  of  series 

can  be  best   discussed  when  we  know  the  voltage  drop  per 

tank,  grouping  them  so  as  to  absorb  either  110  or  220  volts 

per  series  in  one  group. 

(2)  An  anode  sheet  weighs: 

75X50X0.3X7.9X1.1  =  9,776  grams, 

of  which  there  is  immersed  8,888  grams. 
The  17  anodes  per  tank  weigh  altogether: 

9.776X17  =  166.2kg., 
and  the  16  cathodes,  which  are  one-third  as  thick: 

3.259X16  =  521.1  kg. 
In  the  whole  plant,  at  starting,  the  weights  will  be: 

Anode  sheets  166.2X762  =  126,644  kg. 

Cathode  sheets  52.1X762=    39,700    "          (2) 

(3)  Volume  of  liquid  in  tank: 

(1  —  0.1)  X  0.6X2  =     1.08  cubic  meters. 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          449 

Weight  of  solution  per  tank  : 

0.18X1,000X1.1  =  1,188kg. 

Weight  of  dissolved  salts: 

Copperas  =  118.8    " 

Ammonium  sulphate  =    59.4    ' 

Weight  in  the  whole  plant: 

Copperas  =    90.5  tons. 

Ammonium  sulphate  =    45.3   "  (3) 

(4)  At  starting  the  surface  of  the  electrodes  are  6.04  c.m. 
apart,  and  110  amps,  passes  through  each  square  meter  of 
electrode  surface;  therefore,  110X0.375  =  41.25  amps,  pass 
from  each  free  side  of  each  anode  plate  to  the  corresponding 
side  of  a  cathode  plate.  Neglecting  the  small  cross  sectional 
area  of  electrolyte  outside  the  plates,  the  resistance  of  each 
space  would  be 


and  the  drop  of  voltage  across  two  electrodes: 
0.0322X41.25  =  1.33  volts. 

If  we  take  into  account  the  5  c.m.  free  space  at  the  sides 
of  each  electrode,  and  allow  an  equal  amount  as  effective  be- 
neath, the  cross-section  of  the  electrolyte  may  be  taken  as 

(75  +  5)  X  (50  +  10)  =  4,800  c.m2., 
and  the  resistance  between  two  plates: 

20X6.04^4.800=  0.0252  ohms, 
and  the  drop  0.0252X41.25  =  1.04  volts. 

This  value  is  the  more  probable  one  of  the  two. 

At  the  close  of  a  deposition,  neglecting  the  decreased  thick- 
ness of  the  thin  anode  plates,  the  working  distance  is  decreased 
by  one  and  a  half  centimeters,  and  the  voltage  drop  between 
plates  will  then  be: 


x  41.25  -a 


460  METALLURGICAL  CALCULATIONS. 

In  both  cases  the  voltage  drop  in  contacts  and  conductors 
being  0.3  volt,  the  working  voltage  per  tank  will  be: 

Volts. 

At  starting 1.34 

At  end 1.08 

Average 1.21 

The  voltage  needed  at  the  generators  can  only  be  calculated 
when  we  assume  a  plan  of  grouping  the  762  tanks.  If  we  as- 
sume 110  volts  to  be  desired  at  the  generators,  we  could  run 
102  tanks  in  one  series,  which  would  give  7|  series.  If  we 
used  220  volts  at  the  generators  2  series  of  190  cells  and  2  of 
191  would  absorb  at  starting  255  and  256  volts  respectively,  but 
when  in  regular  running,  with  tanks  in  all  stages  of  deposition, 
205  and  206  volts.  This  would  be  a  reasonable  and  practicable 
arrangement.  In  reality,  at  least  one  if  not  two  additional 
tanks  would  be  slipped  into  each  series,  for  at  least  that  num- 
ber would  be  out  of  circuit  continuously,  being  cleaned  and 
made  ready  for  re-starting. 

(5)  The  amperes  per  tank  would  be : 

0.75X0.5X2X16X110  =  1,320 

And  the  area  of  the  main  conductors  in  each  series: 

1,320-^-2  =  660  sq.  m.m.  (5) 

(6)  The   part   of   the   anode   sheet   immersed   weighs   8,888 
grams,  of  which  0.9  is  8,000  grams,  and  if  the  anode  is  an  in- 
termediate one  it  is  corroded  on  both  sides,  and  receives,  there- 
fore, 85  amps,  of  current.     This  current  dissolves,  per  second: 

0.00001036X28X85  =  0  024657  grams. 
And  therefore  the  anode  sheet  will  last 

8,000X0.024657  =  324,000  seconds. 

90  hours.  (6) 

(7)  The  weight  of  deposit  on  both  sides  of  a  cathode  plate  is 

75X50X2X1.5X7.6  =  85,500  grams. 

And  the  time  required  to  deposit  this,  since  it  is  deposited  by 
85  amps.,  is 

85,500  -=-0.024657  =  3,467,600  seconds. 

=  40  days  3  hours.  (8) 


ELECTROMETALLURGY  OF  IRON  AND  STEEL.          451 

(8)   Each  series  requires   1,320  amps,  at  205  volts,  or 
1,320x205-^1,000  =  270.6  kilowatts. 

The  three   series   therefore   require   812   kilowatts,   which  will 
cost 

$25.00X812-^365.25  =  $55.68  per  day. 

An  average  cost  of  power  per  ton  of  iron  refined  of 

$55.68^-25  =  $2.23.  (8) 

The  other  items  of  cost  in  a  well  conducted  refirery  will 
about  equal  this  sum,  making  the  total  cost  of  refining  about 
$4.50  per  ton  of  pure  iron.  With  cheap  soft  steel  used  as  the 
raw  material,  there  is  a  striking  possibility  of  such  a  process 
being  commercially  practicable  for  furnishing  one  of  the  raw 
materials  for  producing  the  finest  qualities  of  steel,  the  other 
raw  material  being  washed  pig  metal  of  standard  quality.  We 
commend  this  possibility  to  the  attention  of  the  makers  of  fine 
steel. 


APPENDIX  TO  PART  II. 

PROBLEMS   FOR  PRACTICE. 

Problem  81. 

A  blast  furnace  will  require  7000  cubic  feet  of  cold  air  sup- 
plied per  minute,  and  assuming  that  it  will  be  provided  with 
5  tuyeres,  and  that  the  pressure  in  the  main  will  be  kept  at  8 
pounds  per  square  inch,  that  the  back  pressure  in  the  furnace 
will  average  2  pounds  per  square  inch,  and  that  the  coefficient 
of  contraction  of  the  jet,  for  a  conical  nozzle,  is  0.92 — 

Required : 

(1)  The  diameter  of  each  nozzle. 

Ans.\    3  inches. 

Problem  82. 

A  blast-furnace  charge  is  composed  of  the  following  ingre- 
dients, in  percentage  composition: 

Ore  Coal                Ash  of  Coal  Flux 

Fe2O3    71.49  fixed  C  90       SiO2     48.00  CaO   52.83 

SiO2      15.73  ash        10       A12O3   41.20  CO2    41.51 

A1203      4.21  Fe2O3     8.00  SiO2     5.66 

CaO        5.00  CaO       2.80 
MgO       3.57 

Per  ton  of  pig  iron  there  is  used  1.5  tons  of  coal.  Pig  iron 
contains  94  per  cent.  Fe,  and  2.8  per  cent.  Si,  99  per  cent,  of  the 
iron  in  the  charge  goes  into  the  pig  iron.  Quantivalent  ratio 
of  the  slag,  calling  both  Si  and  Al  acids,  1.882. 

Required,  per  ton  of  pig  iron  produced: 

(1)  The  weight  of  ore  charged.  Ans. :  1.88  tons 

(2)  a          "        "  flux      "  Ans.: 0.48 tons 

(3)  "         "        "  slagproduced  A ns.\ 0.91  tons 

(4)  The  percentage  composition  of  the  slag. 

452 


APPENDIX.  453 

Ans.:        SiO2 36.95  per  cent. 

APO3 15.56 

CaO 38.79 

MgO 7.40 

FeO..  1.30 


100.00 
Problem  83. 

The  composition  of  the  waste  gases  of  a  blast  furnace  is  by 
weight  (not  by  volume) 

Nitrogen 55 . 40 

Carbonous  oxide 28 . 00 

Carbonic  oxide 16 . 50 

Hydrogen 0. 10 

From  analyses  of  the  charges  and  products  it  is  known  that 
for  every  100  kg.  of  pig-iron  made  79.52  kg.  of  oxygen  enters 
the  gas  from  the  solid  charges  and  15.5  kg.  of  carbon  enters  the 
gases  from  other  sources  than  the  fixed  carbon  of  the  fuel. 
The  fuel  contains  90  per  cent,  fixed  c*arbon;  the  pig-iron  con- 
tains 3  per  cent,  carbon.  In  24  hours  there  is  produced  41,400 
kg.  of  pig-iron;  the  blowing  engine  works  23  hours.  Assume 
blast  dry. 

Required : 

(1)  Per  100  kg.  of  pig  iron,  the  weight  of  fuel  charged. 

Ans. :  114.4  kg. 

(2)  "  "  "       "     the  carbon  burned  at  the  tuyeres. 

Ans.:    87.4kg. 

(3)  "  a  "       "     the  carbon  otherwise  consumed. 

Ans. :  12.6  kg. 

(4)  "  "       "     the  weight  of  the  gases. 

Ans  :  700  kg. 

(5)  "  "  "       "     the  weight  of  the  gases. 

Ans. :  504  kg. 

(6)  The  volume  of  blast  received  per  minute. 

Ans.:  117m3. 

(7)  The  dimensions  of  the  blast  cylinder. 

Ans. :  1  =  1.5  m,  J  =  1.88  m.,  20  r.p.m. 


454  METALLURGICAL  CALCULATIONS. 

Problem  84. 

Four  varieties  of  ore  are  used  in  a  blast-furnace,  containing 
by  analysis: 

A  B  C  D 

Fe 63.25  60.10  64.35  62.35 

SiO2 5.86  4.20  5.30  6.58 

APO3 1.48  1.98  1.96  1.87 

CaO 1.04  0.16  0.90 

MgO 0.75  0.09  0.51 

P 0.019  0.107  0.019  0.015 

The  flux,  fuel  and  pig  iron  contained: 

Flux  Fuel  Pig  Iron 

SiO2 5.46  5.64  Fe    95.24 

A12O3 1.53  3.74  Si       1.40 

CaO 47.00  0.56 

MgO 3.60  0.60 

P 0.010  0.020 

FeS  1.32 

The  ores  are  mixed  in  the  proportions  J  A,  J  B,  f  C,  J  D. 
A  "  round  "  of  fuel  consists  of  12  barrows  of  coke  containing 
520  pounds  per  barrow.  1885  pounds  of  coke  is  used  in  pro- 
ducing a  ."  ton "  of  2300  pounds  of  pig-iron.  Assume  •£$ 
of  the  sulphur  to  go  into  the  slag,  and  TV  of  the  phosphorus 
to  go  into  the  pig-iron. 

Required : 

(1)  A  corrected  fluxing  table,   showing  for  each   substance 
(100  parts)   the  weights  of  ingredients  furnished  to  the  slag, 
the  quanti valence  of  each  element  thus  contributed,  and  the  net 
available  acid  or  basic  quanti  valence  assuming  the  slag  to  be  a 
bi-silicate,  counting  Si  and  Al  as  the  acid  elements  in  the  slag. 

(2)  The  weight  of  each  ingredient  of  the  burden  per  round  of 
coke.  Ans.:  Ore  11,640  lb.,  flux  1090,  coke  6240. 

(3)  The  weight  of  slag  produced  per  2300  Ibs.  of  pig-iron. 

(4)  The  percentage  composition  of  the  slag. 

(5)  The  percentage  of  S  and  P  in  the  pig-iron. 

Problem  85. 

The  flux  used  in  a  blast-furnace  and  the  slag  produced  by  its 
use  have  the  following  compositions: 


APPENDIX.  455 

Flux  Slag 

CaCO3          80.36                       CaO  29.0 

MgCO3         15.75                      MgO  4.8 

SiO2                3.00                       SiO2  53.4 

H20               0.89                      APO3  9.0 

FeO  2.6 

Na2O  1.2 

Per  100  kg.  of  pig  iron  produced  there  is  used  28.7  kg.  of  flux, 
and  there  is  made  50.15  kg.  of  slag. 
Required : 

(1)  The  quantivalent  ratio  of  the  slag. 

Ans.:  2.96 

(2)  How  much  flux  would  be  needed  to  produce  a  uni-silicate 
slag.  Ans. :  105  kg. 

(3)  What  would  be  the  weight  of  this  slag. 

Ans. :  93  kg. 

(4)  What  would  be  its  percentage  composition;  check  the 
quantivalent  ratio  from  this  composition. 

Problem  86. 

A  blast  furnace  works  on  the  following  materials: 

Ore  Coke  Flux  Pig  Iron       Gases 

Fe203    85         Fixed  C  87     CaO     52.50     Fe    94.8     CO   28.41 
SiO2        9         SiO2  6     SiO2       6.25     Si        1.4     CO2  10.85 

APO3      1         APO3         5     CO2      41.25    C        3.8      N2  60.78 
CaO        2.8     H2O  2 

H2O        2.2 


100  kg.  of  coke  and  160  kg.  of  ore  are  used  per  100  kg.  of  pig- 
iron  produced. 

Required:  (1)  The  weight  of  flux  charged,  per  100  of  coke, 
to  make  a  slag  with  the  quantivalent  ratio: 

Quant.  Si  +  Quant.  Al  _ 

Quant.  Ca  Ans.:  39.1kg. 

(2)  The  weight  of  carbon  burned  at  the  tuyeres  per  100  of 
pig-iron.  '   Ans.:  70.1kg. 

(3)  The  weight  of  carbon  consumed  above  the  tuyeres  per  100 
of  pig-iron.  Ans.:  13.1  kg. 


456  METALLURGICAL  CALCULATIONS. 

Problem  87. 

The  blast-furnace  of  preceding  problem  has  7  tuyeres,  each 
5  inches  in  diameter.  Temperature  of  blast  in  blast  main  819°, 
indicated  pressure  on  blast-main  gauge  22.08  Ibs.  per  sq.  inch, 
the  back  pressure  in  the  furnace  in  the  region  of  the  tuyeres  is 
3.69  Ibs.  per  sq.  inch.  The  air  jet  contracts  in  issuing  until  the 
coefficient  of  contraction  of  area  of  the  jet — /JL — is  0.9. 

Required:  (1)  The  actual  temperature  of  the  blast  entering 
the  furnace,  assuming  it  to  be  cooled  only  by  expansion.  As- 
sume the  mean  specific  heat  of  air  under  the  above  conditions 
as  0.40  (oz.  cal.  per  cubic  foot  or  Cal.  per  cubic  meter). 

Ans.:  743° 

(2)  The  volume  of  blast  per  minute  received  by  the  furnace, 
at  standard  conditions  of  measurement.  Ans.:  48,400  ft3. 

Problem  88. 

Taking  the  calculated  data  of  Problems  86  and  87,  and  com- 
bining them,  how  many  tons  of  pig-iron,  at  2240  Ibs.  per  ton, 
are  produced  per  day,  assuming  the  blast  to  be  on  the  furnace 
23  hours  30  minutes?  Ans.:  700  tons. 

Problem  89. 

A  blast  furnace  makes  330,000  kg.  of  pig  iron  in  24  hours. 
The  blowing  engine  runs  1400  minutes  per  day;  3  cylinders, 
double  acting,  2  m.  diameter  by  2  m.  stroke,  piston  rod  0.1  m. 
diameter  and  passing  through  one  end  of  the  cylinder  only;  40 
strokes  per  minute.  Temperature  in  engine  room  27.3°;  assume 
air  dry. 

Charges  per  100  kg.  of  pig  iron:  Ore  155.0,  flux  45.2,  coke 
84.0.  Ore  contains  62  per  cent,  of  Fe,  as  Fe203,  5  per  cent. 
H2O  and  the  rest  SiO2.  The  flux  is  3  per  cent.  SiO2  and  the 
rest  CaCO3.  The  coke  contains  89  per  cent,  fixed  carbon,  1 
per  cent.  H20  and  the  rest  half  SiO2  and  half  A12O3.  The  pig- 
iron  carries  3.5  carbon,  1  silicon,  94.6  per  cent.  iron.  The  waste 
gases  contain  2.3  volumes  of  CO  to  1  volume  of  CO2.  Assume 
all  the  oxygen  in  the  CO  and  CO2  to  come  from  the  blast,  re- 
duction of  Fe2O3  and  SiO2,  and  the  CO2  of  the  flux. 
Required : 

(1)  The  weight  of  carbon  burned  at  the  tuyeres,  per  100  kg.  of 
pig-iron  produced.  Ans. :  67.8  kg. 


APPENDIX.  457 

(2)  The  weight  of  carbon  burned  above  the  tuyeres,  in  direct 
reduction,  per  100  of  pig-iron.  Ans. :  13.5  kg. 

(3)  The  volume  of  blast,  at  standard  conditions,  received  by 
the  furnace  per  minute.  Ans. :  608.5  m3 

(4)  The  same,  per  ton  of  pig-iron.  Ans.:  2582  m3 

(5)  The  efficiency  of  the  blowing  plant,  i.e.,  the  ratio  of  air 
received  to  the  piston  displacement.  Ans. :  88.9  per  cent. 

Problem  90. 

A  blast-furnace  produces  300  metric  tons  of  pig-iron  per  day, 
and  is  charged,  per  100  kg.  of  pig-iron,  with 

156  kg.  iron  ore 
50    "    limestone 
90    "    coke 

The  percentage  compositions  of  these  materials  and  of  the  pig- 
iron  and  gases  produced  are: 


Ore 

Limestone 

Coke 

Pig  Iron 

Gases 

Fe203 

85 

CaO 

51 

.66 

Fixed 

C88 

Fe 

94. 

00 

CO 

26.50 

SiO2 

8 

MgO 

2 

50 

SiO2 

8 

Si 

2. 

10 

CO' 

!  13.25 

APO3 

4 

SiO2 

2 

.50 

FeS 

2 

C 

3. 

75 

N2 

60.25 

H2O 

3 

CO2 

43 

.34 

H20 

2 

s 

0. 

10 

Required : 

(1)  The  volume  of  gas,  measured  dry,  at  0°  and  760  m.m. 
pressure,  produced  per  100  kg.  of  pig-iron  made. 

Ans. :  379  m3 

(2)  The  volume  of  blast,  assumed  dry  and  at  0°  and  760  m.m., 
received  per  100  kg.  of  pig-iron  made.  Ans. :  288  m3 

(3)  The  weight  of  carbon  consumed  by  the  blast,  at  the  tuy- 
eres, per  100  kg.  of  pig-iron  made.  Ans. :  66  kg. 

(4)  The  weight  and  percentage  composition  of  the  slag. 

Ans. :  49  kg. 

Composition:  SiO2  32.6,  APO3 12.4,  CaO  49.3,  MgO  2.4,  FeO  1.2, 
CaS  2.4. 

(5)  The  horse-power  producible  from  one-half  of  the  gases, 
assuming  them  used  in  gas-engines  at  25  per  cent,  net  thermo- 
mechanical  efficiency.  Ans. :  7570  h.p. 

Problem  91. 

A  blast-furnace   makes   300  metric  tons  of  pig-iron   daily, 
producing  per  100  kg.  of  pig  iron  made  464.5  m3  of  gas  of  the 


458  METALLURGICAL  CALCULATIONS. 

following  composition:  CO  24,  CO2  12,  N2  64,  and  using  352.5  m3 
of  blast,  at  700°  C.  The  blowing  engines  consume  1,000  indi- 
cated horse-power,  the  lift  and  pumps  150  h.p.  The  blast  (as- 
sumed dry  at  0°)  is  heated  to  700°  in  fire-brick  stoves,  at  an 
efficiency  of  50%  on  the  heat  of  combustion  of  the  gas  used  in 
them. 

Required : 

(1)  The  proportion  of  the  gas  made  by  the  furnace  required 
to  be  used  by  the  stoves.  Ans. :  46.6  per  cent. 

(2)  The  rest  of  the  gas  being  burned  under  boilers,  and  just 
raising  the  power  required  by  the   furnace,   what  is  the   net 
thermo-mechanical  efficiency  of  the  boilers  and  engines  together? 

Ans. :  3.2  per  cent. 

(3)  If  these  gases  were  used  in  gas  engines  at  an  efficiency 
of  25  per  cent.,  what  surplus  power  would  the  blast-furnace 
produce  above  its  own  requirements?     Ans. :  7830  horse-power. 

Problem  92. 

An  Alabama  blast  furnace  uses  and  produces  materials  of  the 
following  compositions : 


Ore 

Dolomite 

Coke 

Pig  Iron 

Fe203 

72. 

85 

CaO 

30 

.24 

C 

86 

.00 

Fe 

93 

.54 

SiO2 

9.00 

MgO 

20 

.48 

SiO2 

4 

.00 

C 

3 

.50 

A12O3 

3. 

75 

Fe2O3 

0 

.50 

A12O3 

2 

.00 

Si 

2 

.25 

CaO 

0 

75 

A12O3 

0 

.50 

Fe2O3 

2 

.00 

S 

0 

.06 

P2O5 

0. 

90 

SiO2 

2 

.00 

S 

1 

.00 

p 

0 

.65 

SO3 

0. 

25 

CO2 

46 

.28 

H20 

5 

.00 

CO2 

0. 

60 

H20 

11. 

90 

Assume:  Burden  =  2.00  (i.e.,  ore  plus  dolomite  =  twice 
coke).  All  the  iron  in  the  charge  reduced  into  the  pig  iron. 

Sulphur  not  in  pig  iron  goes  into  slag  as  CaS. 

Phosphorus  not  in  pig  iron  goes  into  slag  as  P2O5. 

Slag  to  contain  34.00  per  cent.  SiO2. 

Eighty-nine  per  cent,  of  the  fixed  carbon  of  the  coke  will  be 
consumed  at  the  tuyeres,  by  dry  blast. 

Required:  A  balance  sheet  of  materials  entering  and  leaving 
the  furnace. 


APPENDIX. 


459 


Solution:     [N.B. — Let    %  =  ore    used,    y  =  dolomite,    then 

— ^  =  coke,  and  the  balance  sheet  thus  constructed  leads  easily 
2 

to  a  solution  conforming  to  above  conditions.] 

BALANCE  SHEET  (per  1000  of  pig  iron). 


Charges                             Pig  Iron                   Slag 

Gases 

Ore 

1792 

Fe203 

1305.5                 Fe 

913.9 

0 

391.6 

SiO2 

161.3                 Si 

22.5           SiO2      113.1 

O 

25.7 

oi2o3 

67.2 

A12O3     67.2 

CaO 

13.4 

/Ca             1.5 
\CaO        11.3 

0 

0.6 

P205 

16.1                 P 

6.5           P2O5         1.3 

O 

8.4 

SO3 

4.5                 S 

0.6           S               1.2 

CO2 

10.8 

CO2 

10.8 

H2O 

213.2 

H2O 

213.2 

Flux 

411 

CaO 

124.3 

fCa           13.8 
\  CaO      105.0 

0 

5.5 

MgO 

84.2 

MgO       84.2 

Fe203 

2.1                 Fe 

1.5 

O 

0.6 

A1203 

2.1 

A12O3       2.1 

SiO2 

8.2 

SiO2         8.2 

CO2 

190.2 

CO2 

190.2 

Coke 

1101 

C 

946.9                 C 

35.0 

C 

911.9 

SiO2 

44.0 

SiO2       44.0 

A1203 

22.0 

A12O3     22.0 

Fe203 

22.0                 Fe 

15.4 

O 

6.6 

S 

11.0 

S             11.0 

H2O 

55.1 

H2O 

55.1 

Blast 

4004 

O2 

924. 

O 

924.0 

N2 

3080. 

N  ; 

3080.0 

7308. 

1000.0                     485.9 

5824.2 

Charges 

Slag 

Coke             1000 

SiO2         34.0% 

CaS 

5.7 

Ore               1628 

A12O3        18.8% 

p205 

0.3 

Flux               373 

CaO          23.9% 

inn  2 

Pig  Iron        908 


MgO         17.5% 


460 


METALLURGICAL  CALCULATIONS. 


Problem  93. 

A  foundry  cupola  melts  6,000  kg.  of  pig  iron  per  hour,  using 
480  kg.  of  coke  containing  85  per  cent,  fixed  carbon.  Average 
composition  of  issuing  gases,  by  volume,  N2  75,  CO2  16,  CO  4. 
Some  Fe,  Mn  and  Si  are  oxidized,  forming  the  silicate  Fe5  Mn 
SiO8.  No  carbon  is  oxidized  out  of  or  absorbed  by  the  pig- 
iron.  Assume  the  blast  dry  and  neglect  the  ash  of  the  coke 
and  corrosion  from  the  lining. 

Requirements : 

(1)  The  volume  of  blast  received  by  the  cupola  per  minute. 

Ans. :  59.7  m3 

(2)  The  percentage  loss  in  weight  of  the  pig-iron,  assuming 
it  to  lose  only  Fe,  Mn  and  Si.  Ans. :  4.74  per  cent. 

(3)  The  proportion  of  the  whole  heat  generated  which  is  due 
to  the  oxidation  of  carbon  and  to  the  oxidation  of  the  pig-iron. 

Ans. :  84.5  and  15.5  per  cent. 

Problem  94. 

Dichmann,  in  Stahl  und  Risen,  1  December,  1905,  gives  the 
following  data  respecting  the  Monell  open-hearth  operation,  as 
practised  by  him. 

Charge:    1000  kg.  limestone,  CaO  54.1,  SiO2  1.65,  MgO  0.68. 
3276    "    hematite  ore,  Fe2O3  95.0,  SiO2  3.7. 

These  were  put  on  a  basic  hearth  and  heated  nearly  to  melt- 
ing. There  was  then  run  upon  them  20,300  kg.  of  melted  pig- 
iron.  Starting  at  2.30  P.M.,  the  subsequent  analyses  showed: 


Slag 

FeO      Fe2O3     Mn  P2O5   SiO2 

47.88      6.10  15.22  2.36  17.68 

10.36      3.23  12.67  2.35  23.00 

11.44      2.37  12.04  2.03  22.90 


Metal 

P.M.          C           Si          P        Mn 

2.30 

4.61     0.84     0.15     2.20 

3.00 

4.56     0.19     0.05     0.45 

5.40 

1.47     0.05     0.03     0.63 

5.45 

(819  kg.  more  ore  added) 

6.15 

0.43     0.05     0.03     0.49 

Required: 

(1)  The  weight  of  iron  reduced  into  the  bath  during  the  slag- 
forming  period — 2.30  to  3.00  P.M. 

Ans. :  1219  kg.  =  66%  of  the  iron  in  the  ore. 


APPENDIX.  461 

(2)  The  weight  of  iron  reduced  into  the  bath  during  the  boil  — 
3.00  to  5.40  P.M. 

Ans.  :  705  kg.  =  32.4%  of  the  iron  in  the  ore. 

(3)  The  weight  of  iron  reduced  from  the  additional  ore  added 
—5.40  to  6.15  P  M 

Ans.  :  509  kg.  =  93%  of  the  iron  in  the  ore  added. 

(4)  The  proportion  of  oxidation  produced  by  the  ore  and  by 
the  furnace  gases  in  each  of  these  periods. 

Ans.  :  1st  period  100  and  0. 
2d  "  30  and  70. 
3d  «  73  and  27. 

Problem  95. 

A  Bessemer  converter  is  charged  with  10,000  kg.  of  pig-iron, 
from  which  is  oxidized 

Si  .......................  .  .   2.8  per  cent. 

C  .........................   3.3        "          1/5  to  CO2 

Fe  ........................   1.12      " 

Length  of  blow  15  minutes,  no  free  oxygen  escapes  from  the 
converter.  Temperature  in  engine  room  0°,  barometer  760 
m.m.,  mean  pressure  on  piston  of  blowing  cylinder  during  the 
stroke  1  kg.  per  sq.  centimeter.  No  moisture  in  the  air.  As- 
sume a  coefficient  of  delivery  of  0.60. 

Required  : 

(1)  The  volume  of  blast  received  per  minute. 


(2)  The  dimensions  and  speed  of  the  blowing  cylinder. 

Ans.  :  d  2.4  m.,  /  2  m.,  18  r.p.m. 

(3)  The  horse-power  exerted  by  the  blowing  cylinder. 

Ans.  :  727  h.p. 

(4)  If  the  speed  of  the  blowing  cylinder  is  kept  constant,  but 
the  temperature  in  the  engine  room  becomes  30°  and  saturated 
with  moisture,  how  much  longer  will  the  blow  last? 

Ans.  :  46  seconds. 

Problem  96. 

A  Bessemer  converter  contains  10,000  kg.  of  pig-iron,  from 
which  there  is  oxidized  by  the  blast:  Si  1,  Fe  2,  C  as  CO2  0.7, 
C  as  CO  2.8  per  cent.,  while  no  free  oxygen  or  H2O  vapor  escapes 


462  METALLURGICAL  CALCULATIONS. 

from  the  converter.  The  blowing  engines  have  a  piston  dis- 
placement of  326.5  m3  per  minute.  The  air  in  the  engine  room 
is  at  27.3°,  barometer  741  m.m.,  air  saturated  with  moisture. 
The  blow  lasts  9  minutes  10  seconds,  up  to  the  end  of  the  boil. 

Required : 

(1)  The  coefficient  of  useful  delivery  of  the  blowing  cylinders 
,md  conduits.  Ans. :  66  per  cent. 

(2)  The  average  composition  of  the  gases  produced. 

Ans.:  CO  5.18,  CO2  20.73,  H2  3.38,  N2  70.80  per  cent. 

Problem  97. 

The  gases  from  a  Bessemer  converter  had  the  following  av- 
erage composition  during  the  slag-forming  period  and  the  boil, 
respectively : 

First  3  minutes.          Last  2  minutes. 

O2 0.6  0.1 

CO2 8.4  4.3 

CO 5.2  26.1 

H2 0.6  0.6 

N2 85.2  68.9 

Duration  of  blow  5  minutes.  Weight  of  charge  2500  kg. 
Weight  of  carbon  oxidized  3.9  per  cent,  of  the  weight  of  the 
charge.  Assume  the  blast  constant  per  minute  throughout, 
and  that  the  composition  of  the  slag  formed  from  the  Fe,  Mn 
and  Si,  oxidized  is  Fe  Mn  Si20*.  Neglect  the  moisture  of  the 
blast. 

Required: 

(1)  The  volume  of  the  blast  per  minute.        Ans.:  168.6  m8 

(2)  The  proportion  of  the  total  carbon  oxidized  passing  off 
as  CO2.  Ans. :  30.9  per  cent. 

(3)  The  percentage  loss  in  weight  of  the  charge  from  the 
oxidation  of  its  ingredients.  Ans. :  9.30  per  cent. 

Problem  98. 

A  Bessemer  converter  is  charged  with  8000  kg.  of  pig-iron. 
The  slag-forming  period  lasts  6  '20*,  the  boil  4'  10".  At  the 
end  of  the  boil  800  kg.  of  spiegeleisen  is  added,  and  the  con- 
verter momentarily  turned  up.  The  steel  ingots  obtained  weigh 
7QOO  kg.  Hydrogen  in  gases  comes  from  moisture  in  the  blast. 


APPENDIX.  463 

Temperature  of  air  in  engine  room  30°  C.,  barometer  760  m.m. 
Assume  blast  constant  per  minute.  Blast  cylinder  1.5  m.  in- 
ternal diameter,  2  m.  stroke,  double  acting,  90  strokes  per  min- 
ute, piston  rod  through  both  ends,  diameter  0.15  m.  Ef- 
fective pressure  of  the  blast,  2  atmospheres. 
Analyses : 

Pig  Iron  End  of  End  of  Spiegel  Steel 

1st  period  2d  period 

C            4.00  2.63  0.04  5.00  0.49 

Si           2.00  0.26  0.03  3.00  0.10 

Mn         1.40  0.42  0.01  15.00  0.78 

P            0.05  0.05  0.06  0.12  0.06 

S             0.05  0.05  0.06  0.05 

Fe        92.50  96.59  99.80  76.88  98.52 


In  1st  period 

In  2d  period 

CO 

7.19 

27.45 

CO2 

7.19 

2.60 

O2 

2.60 



H2 

2.81 

2.36 

N2 

80.21 

67.59 

Average  composition  of  gases: 

Final  Slag 
SiO2         48.62 
MnO        21.72 
FeO         29.66 


Required : 

(1)  A  balance  sheet,  showing  constituents  of  the  bath  at  each 
stage  of  the  blow. 

(2)  The  volume  of  blast  per  minute,  at  the  engine  room  tem- 
perature; the  weight  of  blast  per  ton  of  pig-iron  treated. 

(3)  The  relative  volumes  of  gas  issuing  per  minute  in  each 
period. 

(4)  The  proportion  of  the  total  carbon  burned  in  the  blow 
which  is  burned  to  CO  and  CO2. 

(5)  The  proportions  of  the  carbon,  silicon  and  manganese  of 
the   spiegeleisen   lost   during   re-carbonization.     Express   these 
also  as  percentages  of  the  weight  of  steel  produced 

(6)  The  total  weight  of  the  slag ;  the  weight  of  silica  corroded 
from  the  lining. 

(7)  The  volume  efficiency  of  delivery  of  the  blowing  engine. 

(8)  The  net  effective  horse-power  of  the  blowing  cylinder. 


464  METALLURGICAL  CALCULATIONS. 

Solution : 

(1)  BALANCE  SHEET 

Pig  Iron     Loss       End         Loss      End     Spiegel  Loss  Steel 

1st  period  2d  period 

C 
Si 
Mn 
P 
S 
Fe 


320 

120 

200 

197 

3 

40 

4 

39 

160 

140 

20 

18 

2 

24 

18 

8 

112 

80 

32 

31 

1 

120 

59 

62 

4 



4 

0 

4 

1 

0 

5 

4 



4 

0 

4 

0 

0 

4 

7400 

60 

7340 

124 

7216 

615 

49 

7782 

Total  8000  400         7600       370       7230       800       130    7900 

(2)  Volume  received  by  converter  per  minute  at  30°: 

287m3 
Weight  per  ton  of  pig-iron  treated :  436  kg. 

(3)  Relative  volumes  of  gases  per  minute :         1  to  1.20 

(4)  Proportion  of  carbon  burned  to  CO:  75.7  per  cent. 

(5)  Loss  in  per  cent,  of  Spiegel:  in  per  cent,  of  steel 

C  0.5  0.05 

Si  2.3  0.23 

Mn  7.4  0.75 

(6)  Weight  of  slag:  1005  kg.     Loss  of  lining:  112  kg. 

(7)  Efficiency  of  delivery  =  82.5  per  cent. 

(8)  Net  effective  horse-power  =  900;  gross,  1000. 

Problem  99. 

Ten  metric  tons  (10,000  kg.)  of  pig-iron  is  charged  into  a 
basic  lined  (Thomas)  Bessemer  converter,  and  blown  12  min- 
utes. The  lining  is  burnt  dolomite,  composition  practically 
CaO,  MgO.  Analyses  showed: 

Pig  Iron  Metal  at  end  of  blow          Slag 

C            3.05                  C          0.1687            SiO2  5.98 

Mn          0.41                   Mn       0.0973             MnO  1.39 

Si            0.83                   Si                                 CaS  2.49 

S             0.33               •    S           0.0540             P2O5  10.08 

P            1.37                  P          0.0650            FeO  7.14 

Fe        94.01                  Fe      99.6150            Fe203  1.20 

CaO  67.05 

MgO  4.67 


APPENDIX.  465 

Assume  all  the  Fe,  Si  and  Mn  oxidized  to  be  removed  in  the 
first    period,   all    the    carbon   in  the  second,  £  to  CO2  and  £  to 
CO,  and  all  the  P  and  S  in  the  third  period,  and  no  free  oxygen 
to  escape  from  the  converter. 
Requirements  : 

(1)  The  weight  of  slag  produced.  Ans.  :  2975  kg. 

(2)  "         "       «  steel        "  An<r.:9250kg. 

(3)  tt         a        u  lining  corroded  away  during  the  blow. 


(4)  "         "        "  lime  added  during  the  blow. 

Ans.:  1858  kg. 

(5)  The  duration  of  each  period,  assumed  as  sharply  defined. 

Ans.:  2'  27%  6'  60",  2'  43'. 

Problem  100. 

In  a  basic-lined  Bessemer  converter,  there  is  oxidized  during 
the  blow: 

Silicon  .................  0.84  per  cent. 

Carbon.  ...............  3.00        "          1/5  to  CO2 

Phosphorus  .............  1.86 

Manganese  .............  0.55        " 

Iron  ...................  2.80        " 

Bath  weighs  at  starting  8000  kg.,  blow  lasts  18  minutes,  no 
free  oxygen  escapes  from  the  converter.  Effective  pressure  of 
the  blast  1.5  atmospheres,  outside  air  at  0°  and  760  m.m. 

Requirements  : 

(1)  The  volume  of  the  blast  per  minute.  Ans.:  134  m3 

(2)  The  net  horse-power  of  the  blowing  engine. 

Ans.  :  313  h.p. 

(3)  If  the  slag  has  the  formula:  3  Ca4P209.  MnFe5Ca12  Si3O24, 
how  much  lime  must  be  added  during  the  blow  and  what  is  the 
weight  and  percentage  composition  of  the  slag? 

Ans.  :  1075  kg.  CaO 
1905  kg.  slag. 
P205  ........................  17.89  per  cent. 

.     SiO2  ........................  7.56 

CaO  ........................  56.46 

MnO  ..............  .  ........   2.98 

FeO..  ..15.11        a 


PART  III. 
THE  METALS  OTHER  THAN  IRON. 

(NON-FERROUS  METALS.) 


487 


CHAPTER  I. 
THE  METALLURGY  OF  COPPER. 

Roasting  and  Smelting  Copper  Ores. 

The  facts  with  regard  to  the  metallurgy  of  copper  may  be 
found  in  condensed  form,  very  clearly  stated,  in  Schnabel's 
"  Handbook  of  Metallurgy,"  Vol.  I;  they  may  be  found  dis- 
cussed at  greater  length  in  Dr.  Peters'  "  Modern  Copper  Smelt- 
ing "  and  "  Principles  of  Copper  Smelting,"  in  Eissler's  "  Hydro- 
metallurgy  of  Copper,"  and  in  Ulke's  "  Modern  Electrolytic 
Copper  Refining."  *In  the  following  presentation  it  will  be  pos- 
sible to  merely  enumerate  the  different  processes  concerned,  the 
principles  embodied  in  each  and  the  methods  of  calculating 
quantitatively  the  nature  of  the  reactions  involved. 

The  chief  ore  of  copper  is  chalcopyrite,  CuFeS2  which  con- 
tains when  pure  approximately  30  per  cent  iron  and  35  per  cent 
each  of  copper  and  sulphur.  It  is  most  frequently  found  mixed 
with  silica,  SiO2,  as  gangue,  although  various  other  gangue 
materials  are  sometimes  present.  This  mineral  is  fusible,  and 
if  it  were  merely  melted  out  of  the  enclosing  rock,  in  an  atmos- 
phere which  did  not  act  upon  it,  it  would  simply  lose  one-fourth 
of  its  sulphur  and  melt  to  a  fluid  double  sulphide  of  iron  and 
copper,  according  to  the  following  reaction: 

2CuFeS2  =  Cu2S.2FeS  +  S. 

The  melted  sulphide  of  copper  and  iron  resulting  is  called 
"  matte,"  and,  in  fact,  any  mixture  of  Cu2S  and  FeS  in  any  pro- 
portions is  called  matte,  and  in  practice  may  contain  small 
quantities  of  PbS,  ZnS,  BaS,  NiS,  As,  Sb,  Te  and  even  of  Fe3O<, 
which  is  slightly  soluble  in  these  fused  sulphides.  In  the  major- 
ity of  cases  the  matte  is  practically  near  enough  to  a  mixture 
of  Cu2S  and  FeS  to  regard  it  as  containing  only  those  two  sub- 
stances. 

*More  recently,  in  Prof.  H.  O.  Hof man's  splendid  treatise  on  "The 
Metallurgy  of  Copper." 


470  METALLURGICAL  CALCULATIONS. 

Since  the  first  operation  in  the  extraction  of  copper  from  its 
usual  sulphide  ores  is  invariably  the  concentration  to  matte, 
we  will  first  study  the  composition  of  this  substance.  Assum- 
ing matte  to  consist  of  Cu2S  and  FeS  in  varying  proportions, 
we  may  note  that  if  it  is  all  Cu2S  it  contains  2X63.6  =  127.2 
parts  of  copper  to  32  parts  of  sulphur,  or  practically  4  copper 
to  1  sulphur,  or  is  80^£er^ce^itjcapper.  For  -^vjprxJ  0  p^r  _££r\  t 
of  j^^^ec^ujphide  present  there  js^  f.ViPTfor^,  fl  8  per  cent  pf 
co^rje^Jii^theiriatte;  and^xoawssely^Jox^e^ry  1  pex_£ent-Qf 
j]33J4^  ou-lphidc  in 


The  rest  of  the  matte  being  iron  sulphide,  which  contains 
56  parts  iron  to  32  sulphur,  it^aa-  br  r.cicm  tha_LJ:he  per  cent^ 
oMron  ^can  J^e^caIcj^tej^oj^anx^€JLj^nt  ^rf  copper/,  in  fact, 
thp  pej  cfgit  rtf-copper  ^xesboth  that~o£  iron^and^sulphur. 

Illustration:  A  matte  cotrEains  40  per  cent  of  copper.  How- 
much  iron  and  sulphur  does  it  contain?  How  much  copper 
sulphide  and  iron  sulphide? 

40  per  cent  copper  =  40X5/4   =  50  per  cent  Cu2S. 
100—50     =50        "         FeS. 
50  per  cent  FeS  =  50X56  -88  =  31.8     "         Fe. 
100—  (40  +  31.8)=  28.2     "         S. 

We  may  generalize  this  solution  and  say  that  if  X  represents 
the  percentage  of  copper  in  a  matte,  that  the  composition  of  the 
matte  is  as  follows: 

Per  cent  of  copper  =  X 

Per  cent  of  Cu2S  =  1.25  X 

Per  cent  of  FeS  =  100  -  1.25    X 

Per  cent  of  Fe=  (100  -  1.25  X)  56/88  =  63.6  -  0.795  X 

Per  cent  of  S  =  100  -  X  -  Fe  =  36.4  -  0.205  X 

Similarly,  if  Y  represents  the  percentage  of  iron  in  a  matte 
its  composition  is: 

Per  cent  of  Fe  =  Y 

,  Per  cent  of  FeS  =1.57Y 

Per  cent  of  Cu2S  =  100  -  1.57  Y 

Per  cent  of  Cu  =   80  -  1.26  Y 

Per  cent  of  S  =   20  +0.26Y 


THE  METALLURGY  OF  COPPER.  471 

Finally,  if  Z  represents  the  percentage  of  sulphur  in  a  matte 
its  composition  is: 

Per  cent  of  S  =  Z 

Per  cent  of  Fe  =  3.89  Z  -  77.S 

Per  cent  of  Cu  =  177.8  -  4.89  Z 

Per  cent  of  FeS  -  6.11  Z  -  122.3 

Per  cent  of  Cu2S  =  222.3  -  6.11  Z 

The  following  tables   may  then  be  drawn  up,  and  will  be 
found  useful  for  reference: 

Percentages. 

Cu.  Fe.  S.  Cu2S.  FeS. 

80  0.0  20.0  100.0  0.0 


75 

4.0 

21.0 

93.8 

6.2 

70 

8.0 

22.0 

87.5 

12.5 

65 

12.0 

23.0 

81.3 

18.7 

60 

15.9 

24.1 

75.0 

25.0 

55 

19.9 

25.1 

68.8 

31.2 

50 

23.9 

26.1 

62.5 

37.5 

45 

27.9 

27.1 

56.3 

43.7 

40 

31.8 

28.2 

50.0 

50.0 

35  35.8  29.2  43.8  56.2 

J£ 3IL-&_— 30-^- 3?r& 62JL 

25  43.8  31.2  31.3  68.7 


20 

47.7 

32.3 

25.0 

75.0 

15 

51.7 

33.3 

18.8 

81.2 

10 

55.7 

34.3 

12.5 

87.5 

5 

59.7 

35.3 

6.3 

93.7 

0 

63.6 

36.4 

0.0 

100.0 

For  regular  use  in  a  smelting  works  it  is  easy  to  calculate 
a  table  similar  to  the  above  for  every  one  per  cent  of  copper, 
and  keep  it  under  a  glass  cover  for  constant  reference.  Or,  if 
preferred,  a  large  piece  of  cross-section  paper  can  be  taken, 
and  a  diagram  prepared  with  the  percentages  of  copper  as 
abscissas,  from  0  to  80,  and  with  ordinates  running  up  to  100. 
A  line  drawn  from  an  ordinate  36.4  at  abscissa  0  to  ordinate 
20  on  abscissa  80,  will  represent  the  sulphur  content;  a  line 
from  ordinate  63.6  on  abscissa  0  to  ordinate  0  on  abscissa  80, 


472  METALLURGICAL  CALCULATIONS.    ' 

will  represent  the  iron  content;  a  line  from  0  to  ordinate  100 
on  abscissa  80  will  represent  the  Cu2S  content;  and  a  line  from 
ordinate  100  on  abscissa  0  to  ordinate  0  on  abscissa  80  will  rep- 
resent FeS  content. 

It  has  been  already  shown  that  if  a  chalcopyrite  ore  were 
smelted  down,  as  in  a  shaft  furnace,  to  a  matte,  that  about  a 
35  per  cent  matte  is  as  rich  as  could  be  made  by  simple  fusion. 
In  fact,  a  much  poorer  matte  would  usually  result,  because 
chalcopyrite  is  often  accompanied  by  pyrite,  FeS2,  which  on 
simple  heating  becomes  FeS,  and  thus  dilutes  the  matte  still 
further. 

Illustration:  A  chalcopyrite  copper  ore  contains  30  per  cent 
by  weight  of  chalcopyrite  and  25  per  cent  of  iron  pyrite.  What 
grade  of  matte  would  result  from  a  simple  fusion  of  this  ore, 
without  roasting,  in  a  reducing  atmosphere? 

Thirty  per  cent  CuFeS2  contains: 

30  X 160  -J-  368  =  13  per  cent  Cu2S. 

30X176^368  =  14        "         FeS. 

25  per  cent  of  FeS2  will  produce 

25  X   88^-120  =  18       "         FeS. 

Total  matte  =  45        "         of  the  ore. 

Composition  of  matte: 

Cu2S  =  29  per  cent  =  23  per  cent  copper. 
FeS    =  71 

It  is  thus  seen  that  the  presence  of  iron  pyrites  in  the  ore 
tends  to  lower  the  grade  of  the  matte  produced. 

The  essential  principle  of  the  metallurgy  of  copper  is  now 
before  us.  It  is  this  fact:  that  if  some  of  the  sulphur  in  the 
ore'be  first  removed  by  roasting,  and  this  operation  be  followed 
by  smelting,  the  copper  will  first  take  enough  of  the  sulphur 
to  form  Cu2S,  and  then  what  sulphur  is  left  over  will  form  FeS. 
The  amount  of  FeS  which  will  accompany  the  Cu2S  into  the 
matte  is  entirely  a  question  of  how  much  sulphur  is  left  to  com- 
bine with  iron  after  all  the  copper  has  been  satisfied,  and  this 
amount  of  sulphur  can  be  exactly  controlled  by  the  preliminary 
roasting. 

Illustration:  Taking  the  ore  mentioned  in  the  previous  illus- 
tration, containing  30  per  cent  of  chalcopyrite  (=  10.4  per  cent 


THE  METALLURGY  OF  COPPER.  473 

of  copper)  and  25  per  cent  of  iron  pyrites,  what  per  cent  of 
sulphur  does  it  contain,  and  what  would  be  the  quality  of  the 
matte  produced  by  fusion  in  a  reducing  atmosphere  if  the  sulphur 
were  previously  roasted  down  to  J  or  }  or  J  of  its  original 
amount  ? 

Sulphur  in  30  per  cent  chalcopyrite : 

30X128-^-368  =    10.5  percent. 

Sulphur  in  25  per  cent  pyrites: 

25  X   64 -i- 120  =    13.3       " 


Total   =   23.8 

If  the  sulphur  were  reduced  to  J,  J  or  J  its  original  amount 
there  would  remain,  out  of  the  23.8  per  cent,  either  11.9,  5.95 
or  2.97  parts  of  sulphur  per  100  of  original  ore;  i.e.,  per  10.4 
parts  of  the  copper.  The  10.4  of  copper  requires  2.6  of 
sulphur  to  form  Cu2S,  leaving  the  following  amounts  of  sulphur 
in  excess  to  form  FeS : 

Sulphur  left  in  ore 11.90       5.95       2.97 

Sulphur  needed  for  Cu2S 2.60       2.60       2  60 


Sulphur  left  to  form  FeS  9.30       3.35       0.37 

FeS  which  will  be  formed 25.61       9.21       1.02 

Cu2S  formed..  13.00     13.00     13.00 


Matte  formed 38.61      22.21     14.02 

Per  cent  of  Cu  in  matte 27.0       46.8       74.2 

It  is  thus  evident  that  the  degree  of  the  previous  roasting 
determines  absolutely  the  quality  or  grade  of  the  matte  formed 
on  the  subsequent  smelting. 

It  can  be  readily  seen  that  since  partial  roasting  virtually 
oxidizes  a  portion  or  fraction  of  the  ore,  that  it  is  the  practical 
equivalent  of  roasting  if  we  can  get  oxide  ore  to  mix  with  sulphide 
ore.  The  raw  oxide  ore  acts  exactly  as  so  much  roasted  ore, 
and  thus  enriches  the  matte  on  subsequent  smelting. 

Illustration:  What  proportion  of  a  cuprite  (copper  oxide) 
ore,  free  from  sulphur  and  containing  28  per  cent  of  copper,  can 
be  mixed  with  the  chalcopyrite  ore  of  the  previous  illustrations, 
to  produce  a  matte  on  subsequent  reducing  smelting  containing 
50  per  cent  of  copper? 


474  METALLURGICAL  CALCULATIONS. 

One  hundred  parts  of  the  chalcopyrite  ore  contain  10.4  parts 
C.u,  20.8  parts  Fe,  23.8  parts  S. 

The  sulphur,  however,  is  not  all  available  for  making  matte, 
because  one-fourth  of  the  sulphur  in  chalcopyrite  and  one-half 
that  in  pyrite  is  given  off  by  simple  heating.  There  would  be 
lost,  therefore,  and  not  available  for  matte,  sulphur  as  follows: 


S  in  chalcopyrite  10.5  Xi   =   2.6  not  available. 
Sin  pyrite  13.3  Xi    =   6.7     " 


9.3     " 
Available  sulphur  =23.8  -  9.3  =  14.5  parts  per  100  of  ore. 

In  a  50  per  cent  matte  there  is  (see  previous  table)  26.1  per 
cent  of  sulphur;  14.5  parts  of  sulphur  available  for  matte  is  there- 
fore capable  of  producing  14.5-^0261  =  55.5  parts  of  50  per 
cent  matte,  which  would  contain  27.75  parts  of  copper.  But 
there  is  only  10.4  of  copper  in  the  chalcopyrite  ore,  leaving  there- 
fore 16.35  of  copper  to  be  supplied  by  the  cuprite  ore,  and  there- 
fore capable  of  taking  up  16.35-^0.28  =  58  parts  of  the  oxide 
ore  per  100  of  the  sulphide  ore.  A  mixture  of  100  of  the  sulph- 
ide ore  and  58  parts  of  the  oxide  ore  would  therefore  smelt 
down  in  a  reducing  atmosphere  to  a  50  per  cent  matte. 

We  have  carefully  specified  a  reducing  atmosphere  as  the  con- 
dition for  smelting  down  to  produce  the  calculated  results, 
because  under  these  conditions  the  oxygen  of  the  roasted 
ore  or  of  the  raw  oxide  ore  added  cannot  combine  with  the  sul- 
phur of  the  ore  and  go  off  as  SO2,  but  is  taken  up  by  the  carbon 
or  CO  in  the  smelting  furnace.  In  smelting  in  a  shaft  furnace 
with  carbon  as  fuel  these  conditions  exist,  and  the  sulphur  in 
the  charge  can  be  counted  on  as  practically  all  forming  matte. 
If  the  smelting  is  done  in  a  reverberatory  furnace,  with  a  neutral 
atmosphere,  much  sulphur  will  be  removed  from  the  charge  by 
the  oxygen  of  the  roasted  ore  or  of  the  raw  ore  added,  and  a 
richer  matte  results.  The  reactions  of  this  elimination  of  sul- 
phur are  mostly 

Cu2S  +  2CuO    =  4Cu  +  SO2 
Cu2S  +  2Cu2O  =  6Cu  +  SO2 

On  an  average,  a  matte  10  per  cent  richer  in  copper  will  be 
obtained  from  a  given  roasted  ore  by  smelting  it  down  in  a 
reverberatory  furnace,  where  these  reactions  between  sulphide 


THE  METALLURGY  OF  COPPER.  475 

and  oxide  of  copper  can  occur,  than  in  shaft  furnace  smelting 
with  coke,  in  which  the  oxygen  of  the  charge  is  taken  up  by 
the  carbon. 

There  is  a  third  variety  of  smelting  which  is  in  reality  a 
combined  roasting  and  smelting-down  operation.  We  refer  to 
what  is  called  pyritic  smelting,  where  a  sulphide  ore  is  roasted 
in  a  blast  of  air  so  rapidly  that  the  heat  generated  melts  down 
the  charge  and  produces  a  concentrated  matte.  This  operation 
is  done  on  pure  sulphides,  however,  without  preliminary  roast- 
ing, and  will  be  subsequently  discussed  by  itself. 

The  Roasting  Operation. 

When  an  ore  is  roasted  it  loses  some  sulphur  and  takes  up 
oxygen.  The  roasted  ore  will  therefore  have  a  different  weight 
from  the  unroasted  ore.  Thus: 

100  parts  of  Cu2S  become  90.0  parts  of  Cu2O. 

100  parts  of  Cu2S  become         100.0      "      CuO. 

100  parts  of  Cu'S  become 


100  parts  of  FeS2  become  64.5      "      Fe3O4 

100  parts  of  FeS2  become  66.7      "      Fe2O3 

100  parts  of  FeS2  become         126.7      "      FeSO4 

Problem  101. 

A  pyritous  copper  ore  from  Ely,  Vt.   (see  Peters'  "  Modern 
Copper  Smelting,"  p.  133),  contained  before  and  after  roasting, 
in  percentages: 

Before.  After. 

Sulphur  ......................   32.6  7.4 

Copper  .......................     8.2  9.1 

The  condition  of  the  copper  in  the  roasted  sample  was  de- 
termined as 

Per  Cent 
Copper  as  CuSO4  ...........................    1.3 

Copper  as  CuO  ..............................   2.1 

Copperas  Cu2S  .................  .'  ............   5.7 

Required  : 

(1)  The  weight  of  roasted  ore  per  100  of  raw  ore. 

(2)  The  proportion  of  the  sulphur  remo\»ed  by  roasting. 


476  METALLURGICAL  CALCULATIONS. 

(3)  The  grade  of  matte  which  would  be  formed  by  smelting 
down  in  a  reducing  atmosphere  in  a  shaft  furnace. 

(4)  The  grade  of  matte  which  would  be  formed  by  smelting 
down  in  a  neutral  atmosphere  in  a  reverberatory  furnace. 

Solution : 

(1)  Since  no  copper  is  lost,  there  will  be  produced  per  100 
of  raw  ore:    • 

100  x  IT?  =  90-1  of  r°asted  ore.  (1) 

y.  i 

Parts. 

(2)  Sulphur  in  100  of  raw  ore  =32.6 

Sulphur  in  90.1  of  roasted  ore  =  90.1X0.074  =    6.7 

Sulphur  eliminated  =  25 . 9 

Proportion  removed: 

OK   q 

±£J!  =  0.794  =  79.4  per  cent.  (2) 

oZ .  O 

(3)  Per  100  of  roasted  ore  we  have: 

Kg. 

Sulphur  to  form  Cu2S  =  9.1  X 0.25  =    2.3 

Sulphur  to  form  FeS  =  7.4  —  2.3  =5.1 

FeS  formed  =  5.1X88/32  =  14.0 

Cu2S  =  11.4 

Matte  formed  =  25.4 

9  1 

Per  cent  of  copper  ==  7^-7  =  0.358  =  36.8  per  cent.  (3) 

Jo .  4 

(4)  In 

CuSO  +  Cu2S  =  3Cu  +  2SO2 

every  1  part  of  copper  as  CuSO4  reduces  two  parts  of  copper  as 
Cu2S,  and  causes  the  elimination  as  gas  of  1  part  of  sulphur. 
In 

2CuO  +  Cu2S  =  4Cu  +  SO2 

1  part  of  copper  as  GuO  reduces  1  part  of  copper  as  Cu2S,  and 


THE  METALLURGY  OF  COPPER.  477 

causes   the   elimination   of   one-fourth   its   weight   of   sulphur. 
Therefore,  the  sulphur  eliminated  by  the  reactions  given  is: 

Parts. 

By  reaction  of  sulphate  on  sulphide  1.3X1          =     1.3 
By  reaction  of  oxide  on  sulphide  2.1  Xj  =     0.5 


Total  =     1.8 

Sulphur  remaining  to  form  matte  =  7.4  —  1.8  =     5.6 

Sulphur  needed  to  form  Cu2S  =     2.3 

Sulphur  left  to  form  FeS  =  5.6-2.3  =3.3 

FeS  formed  =  3.3  X  88/32  =    9.1 

Cu2S  formed  =11.4 


Matte  formed  =20.5 

9  1 

Per  cent  of  copper  =  =  0.444  =  44.4  per  cent.  (4) 

60 . 0 

In  the  roasting  of  ores  a  large  amount  of  heat  is  set  free, 
since  both  the  sulphur  and  the  metal  are  usually  oxidized. 

Problem  102. 

In  the  ore  of  Problem  101,  taking  the  data  given  for  its 
weight  .before  and  after  roasting,  how  much  heat  was  generated 
in  its  roasting  per  kilogram  of  ore  treated. 

Solution:  Starting  with  100  parts  of  original  ore  it  contained 
32.6  parts  of  sulphur  and  8.2  of  copper.  The  latter  corresponds 
to  2.05  of  sulphur  as  Cu2S,  which  would  mean  8.2  of  sulphur 
altogether  as  chalcopyrite,  leaving  24.4  of  sulphur  present  in 
the  ore  as  iron  pyrites. 

In  100  parts  of  roasted  ore  we  have  5.7- parts  of  copper  as 
Cu2S,  therefore  unchanged,  2.1  parts  changed  to  CuO  and  1.3 
parts  changed  to  CuSO4.  These  contain  sulphur  as  follows: 

Parts. 

Sulphur  for  5.7  Cu  as  Cu2S  =     1.4 

Sulphur  for  1.3  Cu  as  CuSO4  =    0.7 


Sulphur  for  copper  compounds  =    2.1 

Sulphur  for  FeS  =  7.4  -  2. 1  =5.3 

FeS  =  5.3X88/32  =14.6 


478  METALLURGICAL  CALCULATIONS. 

Per  90.1  of  roasted  ore  there  is  present: 

Cuas  CuSO4  =1.17 

Cuas  Cu2S  =5.14 

Cuas  CuO  =  1.89 

Fe  as  FeS  =  9.3 X. 901  =8.38 

Sulphur  =6.67 

We  therefore  have: 

Heat  to  Decompose  Raw  Ore. 

Calories. 

8.20CuasCu2S         =    8.20X160  =1,312 

18.85  Fe  as  sulphide  =  18.85X429  =  8,087 


Sum  =  9,399 

Heat  of  Formation  of  Products. 
5.14CuasCu2S          =    5.14X    160      =       822 
1.17  Cu  as  CuSO4       =     1.17X2,857      =     3,343 
1.89  Cuas  CuO  =     1.89  X    593      =     1,121 

25.93  S  to  SO2  =  25.93X2,164      =  56,112 

9.55  Fe  to  Fe3O4         =    9.55X1,612      =  15,395 
9.30  Fe  to  FeS  =    9.30  X    429      =    3,990 


Sum  =  80,783 

Net  heat  evolved  =  80,783  -  9,399  =  71,384  Calories. 
Per  1  kg.  of  ore  713 

In  the  roasting  of  copper  ore  the  heat  generated  by  the  oxida- 
tion of  the^  ore  is  an  important  item  in  the  heat  required  by  the 
furnace.  In  the  above  case,  for  instance,  each  kilogram  of  pre 
developed  about  one-tenth  as  much  heat  as  a  kilogram  of 
coke,  and  with  proper  arrangements  for  conserving  heat,  such 
as  thick  walls  arid  compact  shape  of  furnace,  combined  with 
regeneration  of  heat  from  the  cooling  ore,  there  is  no  reason 
why  such  an  ore  should  not  be  made  to  roast  itself.  Such  a 
scheme,  when  practicable,  is  highly  economical  in  districts 
where  fuel  is  dear.  A  long-bedded  calciner,  with  its  single 
hearth  and  large  radiating  surface,  cannot  meet  these  condi- 
tions, but  furnaces  having  superposed  hearths  and  arrange- 
ments for  discharging  cold  ore,  cooled  by  the  incoming  air, 
such  as  the  Spence  and  MacDougal  furnaces,  can  work  on 
many  ores  without  using  other  fuel. 


THE  METALLURGY  OF  COPPER.  479 

Problem  103. 

An  improved  Spence  furnace  used  at  Butte,  Mont.,  calcines 
90,000  pounds  of  concentrates  in  24  hours,  of  the  following 
composition : 

Per  Cent. 

Copper 9.8 

Iron 33.8 

Silica 13.3 

Sulphur 41.2 

Four-fifths  of  the  sulphur  is  removed,  and  the  roasted  ore 
is  practically  discharged  cold;  1,500  pounds  of  slack  coal,  cal- 
orific power  6,500,  is  used  per  day.  The  furnace  gases  contain: 

Per  Cent. 

CO2 0.6 

SO2 7.2 

H2O 0.6 

N2 81.3 

O2 10.3 

100.0 

and  escape  into  the  chimney  at  200°  C.  Ore  charged  and  dis^ 
charged  cold.  Furnace  has  an  outside  radiating  surface  of 
5,000  square  feet. 

Required : 

(1)  The  heat  balance  sheet  of  the  furnace. 

(2)  The  proportion  of  the  heat  generated  by  the  roasting  of 
the  ore  and  by  the  combustion  of  fuel. 

(3)  The  heat  radiated  and  conducted  to  the  air  per  square 
foot  of  outside  surface  per  minute. 

Solution : 

(1)  Per  100  of  ore  used  the  unroasted  and  roasted  ore  will 

contain,  respectively: 

Copper '.".' 9.8  9.8 

Iron 33.8  33.8 

Silica 13.3  13.3 

Sulphur .41.2  8.2 

Assuming  that  the  copper  remains  in  the  roasted  ore,  two- 
thirds  as  Cu2S,  one-quarter  as  CuO  and  one-tenth  as  CuSO4, 
while  the  iron  takes  the  rest  of  the  sulphur  to  form  FeS,  the  ex- 


480  METALLURGICAL  CALCULATIONS. 

cess  of  iron  forming  half  Fe2O3  and  half  Fe3O4.     We  have  the 
composition  of  the  roasted  ore  as: 

Cu2S 8.2 

CuO 3.1 

CuSO4... 2.0 

FeS 17.0 

Fe2O3 16.4 

Fe304 15.9 

SiO2 13.3 

75.9 

The  copper  and  iron  existed  in  the  unroasted  ore  virtually 
as  Cu2S,  FeS  and  FeS2,  and  therefore  the  8.2  of  Cu2S  and  the 
17.0  of  FeS  in  the  roasted  ore  can  be  considered  as  unchanged. 
The  actual  change  in  the  roasting  has  been  the  formation  of 
3.1  CuO  and  2.0  CuSO4  from  Cu2S,  and  the  formation  of  16.4 
Fe203  and  15.9  Fe3O4  from  FeS.  If  we,  therefore,  subtract 
from  the  heat  of  formation  of  these  amounts  of  CuO,  CuSO4, 
Fe2O3  and  Fe3O4,  and  the  SO2  formed,  the  heat  required  to 
resolve  the  Cu2S,  FeS  and  FeS2  into  their  constituents,  we 
shall  get  the  net  heat  evolution  in  the  roasting  process.  The 
heats  of  formation  involved  are: 

Molecular  Heat. 

(Cu2,  S)  =    20,300  Calories  =      127  Cal.  per  kg.  of  product. 

(Cu,  O)  =    37,700       «        =     471  " 

(Cu,  S,  O4)  =  181,700        "        =  1,136  " 

(Fe,  S)  =    24,000        "        =     273  " 

(Fe2, 03)  =  195,600        "        =  1,223  " 

(Fe3, 04)  =  270,800        "        =1,167  " 

(S,  O2)  =    69,260        "        =  1,082  " 

We  then  have  the  heat  evolved  and  absorbed  as  follows: 

Evolved.  Calories. 

Formation  of  CuO                       3.1  X    471  =  1,460 

Formation  of  CuSO4                    2.0X1,136  =  2,272 

Formation  of  Fe203                   16.4X1,223  =  20,057 

Formation  of  Fe3O4                   15.9X1,167  =  18,555 

Formation  of  SO2                      66.0X1,082  =  71,412 

Total    -  113,756 


THE  METALLURGY  OF  COPPER.  481 

Absorbed. 

Decomposition  of  Cu2S  4.1  X     127     =  521 

Decomposition  of  FeS  36. IX     273     =        9,855 


Total    =       10,376 
Net  heat  evolution  per  100  of  concentrates: 

113,756-10,376  =  103,380  Calories. 

The  gases  contain  33  pounds  of  sulphur  for  every  100  of  ore 
roasted,  and  the  analysis  shows  there  is  0.072  cubic  foot  of 
SO2  in  each  cubic  foot  of  chimney  gas.  This  weighs  0.072X 
2.88  =  0.2074  ounces,  and  contains  just  half  its  weight  = 
0.1037  ounces  =  0.00674  pounds  of  sulphur.  There  is  therefore 
produced,  per  100  pounds  of  ore,  33  -=-0.00674  =  4,896  cubic 
feet  of  chimney  gas. 

This  contains,  from  its  analysis: 

CO2  .....................  ............      29  cub.  ft. 

SO2  .....  .  .............................    353      "  " 

H2O  ......  t  ..........................      29      "  « 

O2  ..................................    504      " 

N2  .............  ..........  ...........  3,981      "  a 

and  at  200°  C.  carries  heat  up  the  chimney  as  follows: 

Sm  (0  -  200°) 

CO2  29X0.414  =        12  oz.  cal.  per  1° 

SO2          353X0.420  =      148 
H2O  29X0.370  =11        "  « 

°*         4,485X0.308=1,381        «  " 


Sum  =  1,552 

97  Ib.  Cal.  per  1°. 
=  19,400  Ib.  Cal.  per  200°. 

The  fuel    evolves   6,500X1,500  =  9,750,000  pound  Calories 
per  day  =  10,830  pound  Calories  per  100  of  ore  roasted. 

(1)  BALANCE  SHEET  PER  100  OF  ORE  ROASTED. 

Lb.  Calories. 
Evolved  by  the  fuel  ......................    10,830 

Evolved  by  the  roasting  operation  .........  103,380 

Sum  114,210 


482  METALLURGICAL  CALCULATIONS. 

Loss  in  chimney  gases 19,400 

Loss  by  radiation  and  conduction 94,810 


Sum  114,210 

(2)  The  proportion  of  the  total  heat  generated  by  the  ore 
itself  is: 

103'38°       0.90  =  90  per  cent. 


114,210 
by  the  fuel  10 

The  heat  lost  by  radiation  and  conduction  per  day  is: 

Lb.  Calories. 

94,810X900  =   85,329,000 

Per  minute  59,000 

Loss  per  square  foot  of  surface  per 


Problem  104. 

The  Evans-Klepetko  cylindrical  roaster  used  at  Butte,  Mont., 
is  19  feet  high,  18  feet  in  diameter,  and  roasts  80,000  pounds 
of  concentrates  daily  from  35  per  cent  sulphur  down  to  7  per 
cent  (Dr.  Peters).  The  evolution  of  heat  in  the  roasting  opera- 
tion is  sufficient  to  supply  all  the  heat  needed.  The  stirrer 
arms  are  cooled  by  water  circulation,  100  pounds  of  water 
being  used  per  minute,  and  raised  in  temperature  50°  C.  As- 
sume the  evolution  of  heat  to  be  90  per  cent  as  great  as  in  the 
roasting  of  ore  in  Problem  103,  and  the  chimney  loss  to  be 
correspondingly  smaller. 

Required  : 

(1)  The  heat  balance  sheet  per  100  of  ore  roasted. 

(2)  The  loss  by  radiation  and  conduction  per  square  foot  of 
outside  surface  of  the  furnace. 

Solution  : 

(1)  Heat  Evolved. 

Lb.  Calories. 
90  per  cent  of  103,380  =  93,040 


THE  METALLURGY  OF  COPPER.  483 

Heat  Distribution. 

Heat  in  chimney  gases  =  17,460 

Heat  in  cooling  water  =    9,000 

Heat  lost  by  radiation  and  conduction  =  66,580 

Sum  =  93,040      (1) 

(2)  The  outside  surface  consists  of  the  top  and  bottom  and 

sides.     Their  area  is  as  follows: 

Sq.  Feet. 

Sides  18X3.14X19  =    1,074 

Bottom  18X18X0.78  =       253 

Top  =       253 

Total  =     1,580 

Loss  by  radiation  and  conduction  Lb.  Calories. 

Per  100  of  ore  66,580 

Per  day  =  53,264,000 

Per  minute  37,000 

Per  square  foot  surface  per  minute 

37,000  _ 


1,580  "  (2) 

It  is  interesting  to  note,  as  comparing  this  compact  cylindri- 
cal furnace  with  the  rectangular  furnace  of  Problem  103,  that 
although  no  fuel  is  used  and  cooling  water  is  used,  and  radia- 
tion losses  per  square  foot  are  greater  because  of  thinner  walls, 
yet  the  much  smaller  radiating  surface  more  than  counter- 
balanced these  considerations,  and  permitted  the  furnace  to 
run  more  economically. 

Thick  walls  and  minimum  radiating  surface  are  the  requisites 
for  economy  of  fuel  in  general,  and  the  sine  qua  non  for  roasting 
copper  sulphide  ores  by  their  own  self-generated  heat  of  oxida- 
tion. 

Pyritic  Smelting. 

This  method  of  smelting  is  sometimes  called  pyrite  smelting, 
because  usually  practiced  on  ores  rich  in  pyrite;  it  is,  however, 
just  as  applicable  to  ores  rich  in  pyrrhotite  (magnetic  pyrites) 
or  chalcopyrite  (copper  pyrites),  and  therefore  the  more  general 
term  pyritic  smelting  is  really  more  proper  and  descriptive  of 
the  range  of  process. 


-484  METALLURGICAL  CALCULATIONS. 

For  a  full  description  of  pyritic  smelting  we  would  refer  the 
reader  to  the  volume  "  Pyrite  Smelting,"1  a  symposium  of  in- 
formation contributed  by  nearly  forty  metallurgists,  and  edited 
by  T.  A.  Rickard,  or  to  Sticht's  monograph,  "  Ueber  das  Wesen 
des  Pyrit  Schmelzverfahrens,"2  or  to  Dr.  Peters'  "  Principles 
6f  Copper  Smelting,"3  which  contains  a  125-page  chapter  upon  it. 

A  brief  statement  of  the  principles  of  the  process  is  as  follows : 
Given  an  ore  containing  considerable  silica  in  the  free  state 
and  a  good  percentage  of  pyrite,  pyrrhotite  or  chalcopyrite, 
it  is  possible  to  smelt  it  down  in  a  shaft  furnace  to  a  ferrous 
silicate  slag  and  a  matte  by  means  of  cold  blast  and  without 
using  any  carbonaceous  fuel.  Whether  fuel  is  cheap  or  dear  the 
possibility  of  dispensing  with  it  when  smelting  down  certain 
sulphide  ores  is  highly  important ;  yet  it  is  only  within  a  very  few 
years  that  the  principles  involved  have  been  well  enough  under- 
stood to  make  the  process  a  recognized  success.  As  far  as  the 
modus  operandi  is  concerned,  the  ore  is  charged  into  a  shaft  fur- 
nace, water- jacketed  at  the  boshes  and  smelting  zone,  with  enough 
flux  to  furnish  10  to  20  per  cent  of  CaO  or  similar  alkaline  earth 
base  to  the  slag,  and  a  high  pressure  of  blast  is  employed,  such 
as  would  correspond  to  a  high  rate  of  driving  in  ordinary  smelt- 
ing (up  to  3J  pounds  pressure  per  square  inch  is  used).  The 
matte  and  slag  either  collect  in  the  well  of  the  furnace  or  run 
continuously  into  an  external  settler,  where  they  separate  as 
in  ordinary  smelting.  The  heat  necessary  to  run  the  furnace 
is  all  supplied  by  the  oxidation  of  sulphur  and  iron  in  the  fur- 
nace, the  former  escaping  as  SO2  in  the  gases  and  the  latter  as 
silicate  of  FeO  in  the  slag.  The  process  may  be  regarded  as 
a  very  quick  partial  roasting  of  the  ore,  accompanied  by  sim- 
ultaneous formation  of  melted  slag  and  matte,  because  of  the 
high  temperature  generated  by  the  oxidation  itself. 

There  is  hardly  any  process  in  the  whole  of  metallurgy  which 
invites  so  strongly  to  quantitative  calculations,  such  as  we 
are  endeavoring  to  encourage  in  this  treatise;  and  there  are 
fewer  processes  which  present  such  a  lack  of  data  upon  which 
to  base  the  calculations.  Not  only  are  the  physical  and  chemi- 
cal (thermochemical)  data  scarce,  but  the  ordinary  industrial 

1  Engineering  and  Mining  Journal,  New  York,  1905. 

2  Wilhelm  Knapp,  Halle,  1906.     Reprinted  from  "  Metallurgie." 

3  Hill  Publishing  Company,  New  York,  1907. 


THE  METALLURGY  OF  COPPER.  485 

data,  careful  details  of  the  running  of  the  furnaces,  weights 
and  compositions  of  charges  and  products,  analyses  and  tem- 
perature of  gases,  are  very  largely  lacking — most  that  we 
have  of  value  are  those  furnished  as  recently  as  1906  by  Robert 
Sticht,  director  of  the  Mount  Lyell  furnaces  in  Tasmania,  and 
these  apply  only  to  his  particular  furnaces  and  operations. 

FUNDAMENTAL  PRINCIPLES. 

Taking  the  simplest  case,  and  supposing  FeS2  and  SiO2  to  be 
charged  into  a  pyritic  smelting  furnace,  we  know  without  a 
doubt  that  the  FeS2  becomes  approximately  FeS  at  a  red  heat, 
and  the  sulphur  thus  evolved  is  driven  off  at  a  part  of  the 
furnace  where  there  is  no  free  oxygen,  so  remains  in  the  furnace 
gases  as  sulphur  vapor.  At  a  temperature  of  1400°  Sticht 
has  shown  experimentally  that  the  FeS  becomes  something 
like  Fe7S5,  and  it  is  known  that  the  sulphur  in  the  matte  is 
often  less  than  can  correspond  to  FeS.  But  the  actual  tem- 
perature of  combustion  attained  at  the  moment  of  oxidation  in 
pyritic  smelting  is  .higher  even  than  1,400°,  and  it  is  almost 
certain  that  as  FeS  oxidizes,  it  is  heated  so  intensely  that  it 
really  goes  through  two  stages,  first  decomposing  to  Fe2S  and 
then  oxidizing  to  FeO: 

2FeS  =  Fe2S  +  S 
Fe2S  +  SiO2  +  2O2  =  SO2  +  2FeO .  SiO2 

or,  putting  it  all  together: 

2FeS  +  SiO2  +  2O2  =  S  +  SO2  +  2FeO.  SiO2 

If  we  cast  up  the  thermochemistry  of  this  reaction  we  find 
a  very  considerable  evolution  of  heat,  easily  comparable  with 
the  heat  of  oxidation  of  carbon,  and  accounting  for  the  run- 
ning of  the  furnace.  The  thermochemical  analysis  of  the  above 
reaction  is : 

Absorbed.  Calories. 

Decomposition  of  2FeS  48,000 

Evolved. 

Formation  of         2FeO  =    131,400 

SO2  =     69,260 

2FeO.  SiO2   =       8,900 

209,560 


Excess  of  heat  evolved  =    161,560 


486  METALLURGICAL  CALCULATIONS. 

The  FeS  and  SiO2  come  into  the  zone  of  oxidation  already 
heated  to  a  high  temperature  by  their  contact  with  the  rising 
hot  gases.  They  come  to  this  zone,  or  focus,  heated  to  at  least 
1,000°,  and  perhaps  hotter,  before  they  begin  themselves  to 
oxidize.  The  air  conies  in  cool,  and  we  will  not  credit  it  with  any 
sensible  heat  at  the  moment  oxidation  begins. 

THEORETICAL  TEMPERATURE  AT  THE  Focus. 
With  these  data,  and  a  few  reasonable  assumptions,  we  can 
calculate  how  hot  the  focus  will  become  at  its  hottest  point. 
This  calculation  is  similar  to  that  for  the  calorific  intensity  of 
combustion  of  a  fuel,  or  the  theoretical  temperature  before  the 
tuyeres  of  a  blast  furnace.  Making  the  calculation  for  the 
quantities  represented  by  the  equation,  and  assuming 

Heat  in  FeS  at  1000°  melted  200    Calories 

"     "  SiO2  at  1000°  solid  260 

"     "  Slag  (-|-  weight  of  FeO) :  0.27  (t  -  1100)  +300 

o 

"      "  Matte  ft  weight  of  slag) :  0.14  (t-1000)  +200 

we  can  calculate  the  theoretical  temperature  t,  to  which  the 
slag,  matte  and  gases  will  be  raised  by  the  heat  at  hand,  which 
latter  is  the  sensible  heat  in  FeS  and  SiO2  at,  say,  1,000°,  plus 
the  heat  evolved  in  the  reaction,  plus  sensible  heat  in  matte  at 

1000°: 

Calories. 

Heat  in  2FeS  at   1000°:   176X200  =   35,200 

"  SiO2  at  1000°:  60X260  =    15,600 

"       of  the  reaction  =161,560 

in  matte:  80  X  200  =    16,000 


Total  heat  available  =228,360 


Calorific  capacity  of   the  products: 

SO2     =  22.22  (0.36t  +  O.OOOSt2) 
N2       =  170  (0.303t  +  0.000027t2) 
Slag    =  240 [300+  (t-1100)  0.27] 
Matte  =    80  [200+  (t-1000)  0.14] 


Sum    =  0.0113t2  +  135.5t +  5,520         =228,360 
Whence  t  =  1465°C. 


THE  METALLURGY  OF  COPPER.  487 

The  result  of  our  calculation  is  to  show  that  a  temperature 
sufficient  to  run  the  furnace  is  theoretically  obtainable  if  the 
weight  of  slag  made,  carrying  a  given  weight  of  FeO,  is  not  too 
great.  Supposing  the  silica  and  other  slag-forming  ingredients 
are  so  heavy  in  comparison  to  the  amount  of  FeO  formed  by 
oxidation  that  the  slag  contains  only  50  per  cent  of  such  FeO, 
then  the  weight  of  slag  above  would  be  288  instead  of  240,  and 
the  temperature  attained  only  1,365°  instead  of  1,465°.  This 
reduction,  however,  is  getting  perilously  near  the  temperature 
necessary  to  keep  the  furnace  in  operation.  The  formation 
temperatures  of  copper  blast-furnace  slags  are  1,100°  to  1,200°, 
and  an  over-heating  of  at  least  50°  is  necessary,  for  practical 
operation  of  the  furnace.  Suppose  we  put  1,200°  as  the  mini- 
mum theoretical  temperature  which  will  run  the  furnace,  then 
the  maximum  weight  of  slag  can  be  calculated  in  the  above 
solution  for  t,  and  thence  the  minimum  percentage  of  FeO. 

Let  X  be  the  maximum  weight  of  slag,  for  t  a  minimum 
of  1,200°.  Then  we  have: 

Calories. 

Heat  in  SO2  at  1200°  =    19,248 

"     "  N2  at  1200°  =   68,422 

"     "  Slag  at  1200°  330X 

11     "  Matte  at  1200°  =    18,240 

Whence  105,940  +  330X  =  228,360 

and  X=        371kg. 

Since  this  maximum  weight  of  slag  must  contain  the  2FeO  = 
144  kg.  of  FeO,  the  minimum  percentage  of  FeO  produced  by 
oxidation  and  going  into  slag  in  pure  pyritic  smelting,  must  be 

144-^371  =  0.39  =  39  per  cent. 

The  margin  for  working  pyritic  smelting  is  ordinarily  so 
small  that  variations  of  the  temperature  of  the  blast  and  its 
humidity  must  exercise  a  large  influence  in  the  running  of  the 
furnace.  This  coincides  with  experience  as  far  as  the  heating 
of  the  blast  is  concerned.  At  La  Lustre  Smelter,  Santa  Maria 
del  Oro,  Mr.  Koch  says  that  "  a  warm  blast  of  200°  C.  is  a 
sine  qua  non  with  us ;  it  spelled  success ;  cold  blast  meant  failure." 
This  is|  however,  an  extreme  position ;  many  other  metallurgists 
working  other  ores  have  run  entirely  with  cold  blast,  but  there 


488  METALLURGICAL  CALCULATIONS. 

is  no  doubt  that  smelting  is  easier,  faster  and  more  regular 
when  using  warm  blast.  No  one  has  yet  tried  drying  the  blast, 
but  there  can  be  no  doubt  that  under  some  circumstances  this 
would  contribute  greatly  to  the  regularity  of  running  and  would 
increase  the  theoretical  temperature  obtainable  at  the  focus 
or  smelting  zone. 

USE  OF  AUXILIARY  COKE. 

When  the  amount  of  slag-forming  material  in  the  charges 
is  high,  pyritic  smelting  using  cold,  ordinary  blast  may  become 
impossible  for  lack  of  sufficient  calories  to  melt  the  slag  and 
matte.  In  this  case,  which  often  occurs,  some  coke  may  be 
used,  the  combustion  of  which  to  CO2  at  the  focus  increases 
materially  the  heat  available  and  the  temperature.  If  the 
temperature  desired  is,  let  us  say,  1,400°,  carbon  at  1,000°  can 
burn  to  CO2,  yielding  CO2  and  N2  as  products,  and  give  a  large 
surplus  of  heat  to  the  charge  at  this  temperature. 

One  kilogram  of  coke,  containing  0.9  kg.  of  carbon,  will 
form  3.3  kg.  of  CO2  (1.67  cubic  meters)  and  6.35  cubic  meters 
of  N2.  The  heat  generated  is  7,290,  and  adding  in  the  heat  in 
hot  carbon  at  1,000°  (342)  we  have  7,632  Calories  generated. 
But  in  the  products  at  1,400°  we  have: 

Calories. 

CO2 1.67  [0.37 +  0.00022   (1400)]  X 1400  =1582 
N2  6.35  [0.303 +  0.000027  (1400)]  X 1400  =3038 

4620 
Surplus,  to  help  the  fusion 

7632  -  4620  =  3012  Calories. 

Each  kilogram  of  coke  used  will  therefore  melt  down,  at  1400°, 
3012  -T-  330  =  9  kg. 

of  slag,  and  thus  relieves  the  situation  materially.  At  Mount 
Lyell,  0.5,  1.0  and  1.5  per  cent  of  coke  (reckoned  on  the  charge) 
is  found  necessary,  according  to  the  nature  of  the  ore  worked. 
From  this  we  have  increasing  quantities  of  coke  used  up  to 
several  per  cent,  according  to  the  exigencies  of  the  case,  and 
we  pass  by  insensible  gradations  through  partial  pyritic  smelt- 
ing to  ordinary  smelting.  As  the  carbon  is  increased  the  sul- 
phides get  less  and  less  of  the  oxygen  blown  in,  and  therefore 
the  degree  of  oxidation  of  sulphides  is  lower  and  the  concentra- 


THE  METALLURGY  OF  COPPER.  489 

tion  poorer.  The  best  concentration,  using  unroasted  sulphide 
ore,  is  obtained  by  pure  pyritic  smelting,  if  there  are  enough 
sulphides  present  to  generate  the  requisite  heat  and  temperature. 

RATE  OF  SMELTING. 

In  all  that  precedes  it  has  been  assumed  that  the  furnace  was 
driven  fast  enough.  Other  things  being  equal,  the  harder  a 
furnace  is  blown  the  more  material  can  be  put  through  it,  and 
the  smaller  the  heat  losses  by  radiation  and  conduction  when 
expressed  per  unit  of  charge  treated  or  of  product  obtained. 
It  is,  therefore,  possible  to  increase  the  temperature  in  the 
focus  simply  by  harder  blowing,  and  thus  to  decrease  the  amount 
of  auxiliary  coke  needed.  A  similar  effect  is  produced  by  height- 
ening the  urnace  shaft,  since  this  increases  the  regeneration  of 
heat  by  the  charges,  they  coming  to  the  hot  zone  more  intensely 
preheated  by  the  ascending  gases.  The  present  tendency  in 
pyritic  smelting  is  undoubtedly  to  increase  the  rate  of  driving, 
heighten  the  furnace,  and  thus  decrease  the  auxiliary  coke 
needed  and  dispense  with  hot  blast,  which  is  somewhat  of  a 
complication. 

The  possibility  of  smelting  in  any  manner  depends  on  being 
able  to  generate  the  theoretical  temperature  necessary  for  run- 
ning the  furnace,  and  then  keeping  the  rate  of  smelting  per 
minute  per  unit  area  of  the  smelting  zone  as  high  as  prac- 
ticable. This  achieves  two  practical  results,  viz.:  makes  the 
actual  temperature  of  melted-down  slag  and  matte  approximate 
closer  to  the  theoretical  temperature  of  the  focus,  and  gets  the 
largest  tonnage  through  the  furnace.  As  we  gradually  increase 
the  smelting  rate  we  increase  the  temperature  of  the  focus, 
because  of  decreased  radiation  losses;  but,  on  the  other  hand, 
we  tend  to  decrease  the  relative  amount  of  oxidation,  because 
of  the  decreased  time  that  the  charge  is  subjected  to  oxidation; 
there  must  be  a  certain  rate  of  driving  which  will  attain  the 
maximum  of  oxidation,  i.e.,  of  concentration,  and  past  which 
increased  tonnage  is  put  through  at  the  cost  of  decreased  con- 
centration. 

Problem  105. 

W.  H.  Freeland  (Engineering  and  Mining  Journal,  May  2, 
1903),  at  Isabella,  Tenn.,  smelted  Ducktown  pyrrhotite  ore  in 
a  water- jacketed  Herreshoff  furnace,  having  a  cross-sectional 


490  METALLURGICAL  CALCULATIONS. 

area  at  the  tuyeres  of  21.7  square  feet.     The  analyses  of  the 
materials  used  and  the  products  are  as  follows: 

Charges. 

Ore.         Quartz.  Slag.  Coke. 

Cu 2.744           ....  0.73 

Fe 36.519           1.45  39.20  2.30 

S..... 24.848           0.32  1.75  1.58 

SiO2. 18.548         96.79  30.90  8.41 

CaO 7.294           0.23  8.51           

MgO..... 2.672           2.71  .... 

Zn 2.556           2.88           

APO3 0.911           0.32  1.90  3.56 

Mn 0.770           ....  0.85 

O.... 0.38  11.37  1.00 

C.. ....  ....  83.86 

CO2 3.138  

Loss 0.39  

Products. 

Matte.  Flue  Dust.  Slag. 

Cu 20.00  2.20  0.37 

Fe 47.15  30.80  38.84 

S 24.00  16.51  1.74 

SiO2 0.44  23.92  32.60 

CaO....'. 0.10  4.45  8.24 

MgO , 1.38  3.44 

Zn 2.05  2.98  1.54 

A12O3 0.82  1.94  1.50 

Mn 0.53  0.15  0.80 

O 4.91  15.26  10.88 

The  charges  and  products  per  24  hours  and  per  1,000  pounds 

of  ore  used  were 

Charges : 

Ore 68 . 0    tons  1000  Ibs. 

Quartz 5.4       "  80    " 

Slag 9.8       "  145    " 

Coke 2.3       "  34    " 

Products : 

Matte. 8.34  tons  122.65  Ibs. 

Flue  dust 1.75     "  25.71     " 

Slag 63.81     "  938.24    « 


THE  METALLURGY  OF  COPPER.  491 

Blast  applied,  4,500  cubic  feet  displacement  per  minute,  at 
17-ounce  pressure.  Assume  temperature  of  gases  450°  C.,  and 
that  they  contain  no  CO,  SO3  or  free  O2  (no  analyses  are  given). 
Assume  matte  and  slag  issuing  from  the  furnace  at  1,300°  (no 
temperature  given). 

Required : 

(1)  A  balance  sheet  of  everything  entering  and  leaving  the 
furnace. 

(2)  The  volume  efficiency  of  the  blowing  plant. 

(3)  The  heat  generated  per  minute,  per  square  foot  of  cross- 
section,  in  the  focus  of  the  furnace. 

(4)  The  theoretical  temperature  at  the  focus. 

(5)  The  proportion  of  the  heat  generated  in  the  focus  by  the 
combustion  of  carbon  and  by  the  oxidation  of  sulphides. 

(6)  If  hot  blast  were  used  what  should  be  its  temperature 
to  be  able  to  dispense  with  the  coke  charged?     Assume  that 
the  pressure  was  increased  so  as  to  keep  the  delivery  to  the 
furnace  constant  per  minute. 

Per  1000  of  Ore  Smelted. 

Charges.  Matte.  Flue  Duct.  Slag.       Gases. 

Ore  (1000). 

Cu ,. 

Fe 

S 248.48         29.44         4.24       16.33       98.47 

SiO2 

CaO...... 

MgO 

Zn 

APCP 

Mn 

CO2 31.38  31.38 

Quartz  (80). 

Fe 1.16  1.16  

S 0.26  0.26 

SiO2 77.43  77.43 

CaO 0.18  0.18  

APO- 0.26  0.26  

IPO...  0.71  0.71 


27.44 

24.53 

0.57 

2.34 

365.19 

57.83 

7.92 

299.44 

248.48 

29.44 

4.24 

16.33 

185.48 

0.54 

6.15 

178.79 

72.94 

0.12 

1.14 

71.68 

26.72 

.... 

0.35 

26.37 

25.56 

2.51 

0.77 

22.28 

9.11 

1.00 

0.50 

7.61 

7.70 

0.64 

0.14 

6.92 

31.38 

492  METALLURGICAL  CALCULATIONS 

Charges.  Matte.  Flue  Dust.   Slag.       Gases. 

Slags  (145). 

Cu 1.06  1.06 

Fe ..56.84  56.84         .... 

S 2.54  2.54 

SiO2 44.81  44.81          

CaO 12.34  12.34 

MgO 3.93 3.93 

Zn 4.18  4.18         

APO3 2.76  2.76         

Mn 1.23  1.23         

O 15.31  6.04         3.93         5.34 

Coke  (34). 

Fe 0.78  0.78         .... 

S 0.54  0.54 

SiO2 2.86  ....          ....         2.86 

C . 28.51  28.51 

APO3 1.21  1.21          

0 0.10  0.10 

Blast  (1,191). 

O2 274.91  97.92     176.99 

N2...                ..916.37  916.37 


2450   122.65   25.71  946.16  1355.77 

Notes  on  the  Balance  Sheet.         • 

The  oxygen  of  the  blast  goes  as  SO2  and  CO2  into  the  gases 
and  as  oxides  into  the  slag.  The  amount  required  for  the 
gases  is 

00 

Oin  SO2  =SX  =  100.96 


Oin  CO2  =CX  7^  =    76.03 

1.40 


176.99 


0  for  Fe  =  330.68  X  =    94.48 

OD 


THE  METALLURGY  OF  COPPER.  493 

O  for  Zn  =  26.46  X  ^  -       6.51 

O  for  Mn  =    8.15X-^r  =       2.37 


O  in  slag  =  103.36 
Contributed  by  charges  =       5 . 44 

Contributed  by  blast  =     97 . 92 

To  burn  sulphides  and  carbon  at  the  focus  =  176.99 

Total  from  blast  =  274.91 

(2)  The  furnace  receives  1,191  pounds  of  blast  per  1,000  of 
ore  smelted.     This  represents  at  0°  C. : 

1191X16 

=  14,738  cubic  feet. 


1,293 

Per  2000  Ibs.  ore        =        29,475  " 

Per  68  tons  ore  =  2,004,300  "         "    per  day 

=       83,510  "         "      "  hour. 

1,400  "         "      "  minute. 

At  50°  C.  1,477  "         "      " 

Blower  displacement  =          4,500 


a 


1  477 

Efficiency  of  blower  =    '         =  0.328  =  32.8  per  cent.      (2) 
4, out) 

[It  is  high  time  that  practical  furnace  men  should  stop  giv- 
ing the  mechanical  displacement  of  their  blower  as  the  amount 
of  air  received  by  their  furnace.  They  still  keep  doing  that, 
although  the  furnace  receives  only  from  85  per  cent  down 
to  25  per  cent  of  the  given  volume.  What  per  cent  of  the 
piston  displacement  the  furnace  is  actually  receiving  is  a  highly 
important  datum,  but  is  more  often  unknown  than  known  to 
the  practical  men.  It  can  usually  only  be  found  by  calculation, 
as  is  illustrated  in  the  preceding  case.  We  urge  upon  practical 
furnace  managers,  for  their  own  information  and  use,  to  cease 
being  satisfied  with  piston  displacement,  and  to  get  deeper  into 
the  real  inwardness  of  their  furnace  processes  by  calculating 


494  METALLURGICAL  CALCULATIONS. 

the   air   actually   received   by  their   furnaces.     The  two    things 
are  enough  different  to  make  it  always  "  worth  while."] 

(3)  To  calculate  the  heat  generated  at  the  focus  we  will 
assume  that  all  the  fixed  carbon  of  the  coke  is  there  burned  to 
CO2,  and  that  the  rest  of  the  oxygen  blown  in  produces  the 
reaction  characteristic  of  pure  pyritic  smelting.  We  have 
treated  per  minute 

68X2000 

- — •  =  94.4  Ibs.  of  ore. 
1440 

and  the  carbon  burned  per  1000  of  ore  is  28.51,  generating 

28.51X8100  =  230,930  Ib.  Calories, 
this  absorbs 

28.51X8/3  =  76.03  Ibs.  of  oxygen. 

leaving  274.91  -  76.03  =  198.88  Ibs.  of  oxygen  to  oxidize  sul- 
phides. 

Since,  now,  2O2  generates  by  the  pyritic  smelting  reaction 
161,560  calories,  we  have  generated  per  pound  of  oxygen  thus 
used: 

161,560-f-64  =  2,524  Ib.  Calories. 

and  per  1000  of  ore  smelted  we  will  have 

2,524  X198.88  =  502,000  Ib.  Calories, 
total  heat  generated 

502,000  +  230,930  =  732,930  Ib.  Calories. 
Since  this  is  per  1000  of  ore  smelted,  per  minute  we  have 

732,930-;- 1000X94.4  =  77,640  Ib.  Calories, 
and  per  square  foot  of  smelting  zone  area 

77,640-^21.7  =  3,576  Ib.  Calories.  (3) 

This  last  figure  is  extremely  useful  in  determining  the  rela- 
tive activity  of  different  furnaces,  and  in  most  cases  will  be  a 
reliable  index  of  the  rate  at  which  the  furnace  is  capable  of 
being  driven. 


THE  METALLURGY  OF  COPPER.  495 

(4)  Taking  as  the  basis  of  calculation  1,000  pounds  of  ore 
smelted,  there  is  generated  at  the  focus  732,930  pound  Calories, 
there  is  used  1,190  pounds  of  blast,  and  there  arrives  at  the 
focus  all  of  the  charges  except  the  flue  dust,  CO2  of  ore,  H2O  of 
quartz,  and  approximately  one-quarter  of  the  sulphur.  We 
therefore,  have  arriving  at  the  focus  about 

28.51  Ibs.  of  fixed  carbon. 
526.50  Ibs.  of  sulphides. 
621.20  Ibs.  of  inert  slag-forming  material. 

These,  assuming  them  to  reach  the  focus  at  1,000°,  would 
bring  back  into  it  the  following  amounts  of  heat: 

Carbon  28.51  XT380  =     10,834  Ib.  Calories. 

Sulphides,  melted          526.50  X200  =  105,300    " 
Slag-forming  material  621.20X174  =  108,089    " 

224,223    " 

Heat  generated  at  the  focus  =  732,930  "  " 


Total  heat  available  at  the  focus    =957,153   " 

Letting  t  be  the  theoretical  temperature  at  the  focus,  then 
we  have 

Heat  in  Slag      946  [300  +  (t  -  1 100)  0.27] 
«      "  Matte    123  [200 +  (t- 1000)  0.14] 
"      "  S  vapor  76  [  1 79  +  (t  -  445)  0.11] 
"      "  SO2         35[0.36t  +  0.0003t2] 
"       "CO2         53[0.37t  +  0.00022t2] 
«       «  N2         727[0.303t  +  0.000027t2] 

[The  76  pounds  of  sulphur  vaporized  at  the  focus  is  the 
total  sulphur  charged,  less  the  25  pounds  assumed  as  vaporized 
above  the  focus,  and  less  the  100.96  pounds  oxidized  at  the 
focus;  the  expression  is  the  total  heat  in  1  pound  of  sulphur 
vapor  at  t,  in  pound-Calories.  The  70  of  SO2  is  the  weight  of 
sulphur  dioxide  formed,  in  pounds,  divided  by  2.88;  the  53  of 
CO2  is  the  weight  of  CO2  formed,  in  pounds,  divided  by  1.98; 
the  727  of  N2  is  the  weight  of  N2  divided  by  1.26.  To  be  strictly 
logical  the  weights  of  SO2,  CO2  and  N2  should  be  first  multiplied 
by  10  to  get  ounces,  and  the  resulting  expression  in  ounce- 


496  METALLURGICAL  CALCULATIONS. 

calories  divided  by  16  to  get  pound-Calories.     We  have  simply 
dispensed  with  these  two  numerical  operations.] 

The  sum  of  the  calorific  capacity  of  the  products  at  t°  is: 

20,098  +  530.5t  +  0.0425t2  =  957,153 
whence  t  =  1569°  (4) 

It  will  be  noted  that  the  temperature  is  much  higher  than 
could  have  been  attained  by  attempting  to  smelt  this  mixture 
without  coke.  If  the  coke  were  omitted  from  the  charge  we 
would  be  without  the  28.51  pounds  of  fixed  carbon  and  some 
5.5  pounds  of  slag-forming  material;  the  products  would  be 
without  the  CO2  and  the  N2  corresponding  to  it,  and  the  heat 
available  would  be  decreased  230,930  pound-Calories  from  the 
absence  of  carbon.  If  these  corrections  are  made,  and  it  is 
still  assumed  that  the  materials  come  down  into  the  smelting 
zone  at  1,000°,  the  theoretically  calculated  temperature  is 

t  =  1447° 

While  this  temperature  is  theoretically  sufficient,  yet  varia- 
tions in  ore  quality  and  temperature  and  humidity  of  the  air 
would  make  running  under  these "  conditions  much  more  pre- 
carious than  with  the  coke. 

(5)  This   has   already  been    calculated.     We   found   230,930 
pound-Calories  to  be  due  to  the  formation  of  CO2  and  502,000 
to  the  oxidation  of  sulphides,  making  of  the  total 

31  per  cent  from  oxidation  of  carbon 

69  per  cent  from  oxidation  of  sulphides.  (5) 

(6)  If  coke  were  dispensed  with,  the  theoretical  temperature 
previously  attained,   1,569°,  might  be  reached  if  the  air  used 
were  heated.     With  cold  blast  we  have,  under  (4),  calculated  a 
theoretical  temperature  of  1,447°.     Under  these  conditions  the 
heat  available  was  altogether  725,266  pound-Calories.     But  to 
heat  the  products,  whose  heat  capacity  was 

20,083  + 451. 7t  +  0.0247t2 

to  t  =  1,569°  would  require  (evaluating)  789,600  pound-Cal- 
ories, which  is  64,334  pound-Calories  more  than  are  available. 
To  supply  this  difference  by  heating  the  blast  we  reckon  first 


THE  METALLURGY  OF  COPPER.  497 

the  amount  of  blast  per  1,000  of  ore  smelted,  which  will  be  862 
pounds  and  its  heat  capacity: 

862   (0.303t  +  0.000027t2) 


1.293 

making  this  equal  to  the  heat  to  be  supplied,  64,334  Ib.  Cal.  we 
have 

202t  +  0.01805t2  =  64,334 
whence 

t  =  313°.  (6) 

Problem  106. 

At  Mount  Lyell,  Tasmania,  R.  Sticht  analyzed  the  gases  at 
different  depths  of  the  shaft  in  a  pyritic  smelting  furnace,  and 
found  them  of  nearly  uniform  composition  for  6  feet  down, 
leaving  out  the  sulphur  vapor,  which  condensed  in  taking  the 
samples.  At  6  to  7  feet  below  the  top,  close  to  the  focus  of  the 
furnace,  the  mean  of  5  analyses  gave  by  volume 

H2  0.00 

SO3  0.00 

SO2  7.90 

CO2  3.56 

CO  0.00 

O2  0.88 

N2  (difference)  87.66 

The  CO2  comes  from  coke,  which  it  was  stated  was  used  to  the 
extent  of  1.5  per  cent  of  the  charge.  The  slag  contained  53 
per  cent  FeO,  30  per  cent  SiO2. 

Required : 

(1)  The   percentage   of   the   heat  generated   in   the   furnace 
coming  from  the  oxidation  of  sulphides  and  the  combustion  of 
carbon. 

(2)  The  heat  developed  in  the  furnace  per  unit  of  slag  formed. 

(3)  The  volume  of  blast  per  1,000  kg.  of  slag  formed. 

Solution : 

(1)  We  will  assume  the  reaction  of  oxidation  of  sulphides 
to  be 


498  METALLURGICAL  CALCULATIONS. 

Fe2S  +  202  =  S02  +  2FeO 

and,  in  fact,  the  above  analyses  are  our  chief  justification  for 
so  doing.  The  air  blown  in  is  approximately  20.8  per  cent  of 
oxygen  by  volume,  and  according  to  the  above  equation  the 
oxygen  going  to  form  FeO  is  equal  to  that  going  to  form  SO2. 
We  have  then  for  the  oxygen  used 

O2  to  form  7.90  SO2  ..................  7.90 

O2  to  form  -  FeO  .................  7  .  90 

O2  to  form  3.56  CO2  ..................  3.56 

O2  to  form  0.88  O2  ...................  0.88 

Sum     20.24 


°0  8 
O2  corresponding  to  N2  87.66  X  =  23-°° 


There  is  thus  a  small  deficit  of  oxygen,  but  if  the  sulphide 
oxidizing  were  assumed  to  be  FeS  the  reaction  would  be 

2FeS  +  3O2  =  2SO2  +  2FeO 
and  the  oxygen  accounted  for  from  the  gases  would  be  only 

O2  to  form  7.90  SO2  ...............  ...  7.90 

O2  to  form-    -FeO  ...........  ...'...  3.95 

O2  to  form  3.  56  CO2    ................  3.56 

O2  to  form  0.88  O2  ...................  0.88 

16.29 

which  barely  accounts  for  two-thirds  of  the  oxygen  which 
must  have  accompanied  the  nitrogen. 

The  conclusion  must  therefore  be  that  the  sulphide  which  is 
being  oxidized  is  Fe2S,  and  not  FeS,  for  otherwise  the  gas 
analyses  would  show  impossible  conditions. 

The  oxidation  of  the  sulphides  gives  for  each  O2  thus  used 
161,560^-2  =  80,780  Calories,  and  the  oxidation  of  C  to  CO2 
97,200.  From  the  above  analyses  we  infer  that  for  3.56  volumes 
of  oxygen  used  for  CO2  15.80  volumes  were  used  in  oxidizing 
sulphides.  The  relative  amounts  of  heat  thus  generated  are 
therefore  : 


THE  METALLURGY  OF  COPPER.  499 

By  carbon       97,200  X  3.56  =      346,000  =  21.3  per  cent. 
By  sulphides  80,780X15.80  =  1,276,300  =  78.7    "       "    (1) 

(2)  The  production  by  oxidation  of  2FeO  (144  kilograms) 
is  accompanied  by  the  production  of  SO2  (64  kilograms)  and 
the  evolution  of  161,560  Calories.  The  slag  being  53  per  cent 
FeO,  the  slag  formed  by  the  reaction  is  144-^-0.53  =  272  kg. 
The  64  kg.  of  SO2  would  be,  in  volume,  22.22  cubic  meters,  and 
would  be  accompanied  in  the  gases  by 


=  10m3  of  CO2. 


which  contains     10X0.54  =  5.4  kg.  of  C. 
whose  heat  of  oxidation  to  CO2  is 

5.4X8100=     43,740  Cal. 
but,  heat  from  sulphides   =  161,560     " 


total  heat  available        205,300     " 
per  kg.  of  slag  205,300  *  272  =  755  Cal.  (2) 

Per  unit  weight  of  charge  this  heat  would  be  slightly  lower. 
(3)  The  oxygen  used  per  272  kg.  of  slag  is 

for  SO2  22.22m3 

"    FeO  22.22    " 

"    CO2  10.00    " 


Total  54.44    ' 

per  kg.  of  slag  0.20    " 

"    1000kg.  of  slag  200. 

Volume  of  air  per  1000  kg.  of  slag  made 

200-7-0.208  =  960  cubic  meters  (3) 


SMELTING  OF  COPPER  ORES. 

The  smelting  down  to  matte  is  done  either  in  reverberatory 
furnaces  or  in  shaft  furnaces.  The  charges  are  usually  com- 
posed of  roasted  ore,  roasted  matte  or  speiss,  mixed  with  un- 
roasted  sulphides,  usually  concentrates,  and  with  siliceous  rock 


5CO  METALLURGICAL  CALCULATIONS. 

or  limestone  as  flux.  The  important  reaction  during  the  smelt- 
ing down  is  the  formation  of  Cu2S  by  the  copper  present,  the 
formation  of  other  sulphides,  mostly  FeS,  by  the  larger  part 
of  the  sulphur  left  over,  and  the  slagging  of  the  elements  which 
do  not  enter  the  matte.  If  much  lead  is  present  metallic  lead 
will  separate  out,  carrying  the  bulk  of  the  precious  metals,  but 
this  phenomenon  will  be  considered  under  the  metallurgy  of 
lead.  The  important  subjects  for  calculation  are:  the  propor- 
tions of  roasted  and  unroasted  materials  to  be  used,  and  the 
proportions  of  flux  to  use  to  make  a  satisfactory  slag. 

Some  data  of  importance  for  calculations  on  copper  smelting 
have  been  determined  in  the  author's  laboratory  by  Prof.  Walter 
S.  Landis. 

A  copper  matte  containing  47.3  copper,  26.2  iron  and  23.6 
sulphur,  was  found  to  have  the  following  thermo-physical 
characteristics : 

Melting  point  1,000°  C. 

Mean  specific  heat,  0  - 1  =  Sm  =  0.21104  —  0.0000366t 

Actual  specific  heat,  at  t  =  S  =  0.21104  —  0.0000732t 
Heat  content,  solid,  at  1,000°  174  Cal. 

Heat  content,  liquid,  at  1,000°  204  Cal. 

Latent  heat  of  fusion,  at  1,000°  30     " 

Specific  heat,  at  1,000°  =        0.138 

Heat  of  formation  from  Cu2S  and  FeS  not  satisfactorily  de- 
termined. 

A  copper  blast  furnace  slag  containing  35.5  SiO2,  39.7  FeO, 
1.0  MnO,  11.4  CaO,  2.7  MgO,  9.2  A12O3,  0.42  Cu,  0.42  S,  was 
found  to  have  the  following  characteristics: 

Melting  point  1,114°  C. 

Mean  specific  heat,  0  - 1  =  Sm  =  0.20185  -f  0.0000302t 

Actual  specific  heat,  at  t  =  S     =  0.20185  +0.0000604t 
Heat  content,  solid,  at  1,114°  262  Cal. 

Heat  content,  liquid,  at  1,114°  302     " 

Latent  heat  of  fusion,  at  1,114°  40     " 

Specific  heat,  at  1,114°  0.269     " 

Heat  of  formation  from  its  constituent  oxides  =  133  Calories 
per  kg.  of  slag. 


THE  METALLURGY  OF  COPPER.  501 

In  the  case  of  the  slag,  its  melting  point  and  latent  heat  of 
fusion  are  not  nearly  as  sharply  denned  as  those  of  the  matte, 
since  the  matte  melts  sharply,  but  the  slag  goes  through  a  pasty 
or  viscous  stage.  The  values  given  are  the  best  approxima- 
tions which  could  be  gotten  from  the  experiments.  As  regards 
heat  of  formation,  the  constituent  oxides  were  carefully  weighed, 
mixed  with  a  known  weight  of  carbon,  and  ignited  in  a  Berthelot 
bomb  calorimeter.  Several  experiments  gave  satisfactory  con- 
cordant results,  so  that  the  heat  of  combination  of  the  con- 
stituents of  this  slag  may  be  taken  as  not  far  from  133  Calories 
per  kilo  of  slag.  By  following  this  scheme  it  is  possible  to 
measure  experimentally  the  heat  of  formation  of  any  slag 
whose  analysis  is  known. 

I.  Reverberatory  Smelting. 

In  the  reverberatory  furnace  the  atmosphere,  is  in  no  case 
strongly  reducing;  it  may  be  varied  from  weakly  reducing  to 
strongly  oxidizing,  and  in  the  smelting  operation  usually  aver- 
ages about  neutral.  The  consequence  of  this  is  that  copper 
oxides  or  sulphate  in  the  charge  are  not  deoxidized  by  carbon, 
as  in  the  shaft  furnace,  nor  is  Fe2O3  reduced  to  FeO  by  carbon, 
but  the  oxygen  thus  contained  in  the  charge  largely  goes  off 
with  sulphur  as  SO2,  thus  decreasing  the  amount  of  sulphur 
left  to  form  matte,  and  therefore  increasing  the  percentage  of 
copper  in  the  matte  formed.  The  reactions  are  mainly: 

2CuO  +  2FeS  +  2Fe203  +  6Si02  =  SO2  +  Cu2S  +  GFeSiO3. 

The  slag  FeSiO3  is,  however,  heavy  and  viscous;  it  would 
contain  45  per  cent  of  silica,  but  runs  better  and  gives  a  cleaner 
separation  from  matte  if  some  lime  is  added  to  it.  The  re- 
placement of  10  per  cent  of  FeO  in  this  silicate  by  10  per  cent 
of  CaO  lowers  its  melting  point  from  1,110°  C.  to  1,010°,  and 
therefore  makes  a  slag  that  is  much  easier  to  keep  fluid  and  of 
greater  fluidity  at  any  given  furnace  temperature. 

Problem  107. 

Peters  (Modern  Copper  Smelting,  p.  446)  gives  the  com- 
position of  the  average  mixture  smelted  in  the  reverberatory 
furnaces  at  Argo,  Col.,  as 


602  METALLURGICAL  CALCULATIONS. 

Per  Cent. 

SiO2 33.9 

Iron 10.8 

BaSO4 .15.5 

APO3 5.6 

CaCO3 8.5 

MgCO3 5.8 

ZnO 6.1 

Copper :  . 2.0 

Sulphur 5.1 

Oxygen 6.4 


99.7 

The  furnace  smelts  50  tons  of  this  mixture  (charged  hot,  at 
350°  C.)  in  24  hours,  using  13.5  tons  of  bituminous  coal  and 
producing  a  matte  with  40  per  cent  of  copper.  Outside  dimen- 
sions of  furnace  20  x  40  x  6  feet.  Area  of  stack,  fire-box  and 
hearth  16,  32.5  and  481  square  feet,  respectively.  Temperature 
of  stack  gases  1,000°  C.  Composition  of  the  coal:  moisture 
1.40  per  cent,  fixed  carbon  54.90,  volatile  matter  32.90,  ash 
10.80  per  cent;  assume  10  per  cent  more  air  used  than  necessary 
for  theoretical  combustion.  Temperature  of  slag  and  matte 
1,200°. 

Required : 

(1)  The   weight   of  matte   produced,    assuming   the   slag   to 
carry  0.2  per  cent  of  copper  (as  intermingled  matte),  and  the 
matte  40  per  cent. 

(2)  The  loss  of  copper  in   the  slag,   expressed  in  per  cent  of 
the  total  copper  present. 

(3)  The  percentage  of  the  calorific  power  of  the  fuel  exist- 
ing in  the  stack  gases,  the  slag  and  the  matte. 

(4)  The   heat   lost   by   radiation   and   conduction  in  pound- 
Calories  per  minute  per  square  foot  of  furnace  surface. 

(5)  The  velocity  of  the  hot  gases  at  the  base  of  the  stack. 

(6)  The  horse-power  theoretically  obtainable  by  passing  the 
hot  gases  through  a  boiler  which  reduces  their  temperature  to 
200°   C.,   and  which,   together  with  the  steam  engine,  gives  a 
total  thermomechanical  efficiency  of  7.5  per  cent  upon  the  heat 
furnished  to  (entering)  the  boiler. 


THE  METALLURGY  OF  COPPER.  503 

Solution : 

(1)  The  entire  matte  formed,  that  free  plus  that  intermingled 
with  the  slag,  will  be  per  100  of  ore  mixture 

2.0^0.40  =  5.0  pounds. 

-   Pounds. 

FeS  in  this  matte  =  5.0-  2.5  .  =     2.5 

Fe  in  this  matte  =  2.5X56/88  =      1.6 

Fe  going  to  slag  as  FeO  =  10.8-  1.6  =      9.2 

FeOinslag  =  9.2X72/56  =    11.8 

Constituents  of  slag: 

SiO2  =  33.9 

FeO  =  11.8 

BaO  =  15.5  X  153/233  =  10.2 

APO3  =  5.6 

CaO  =  8.5X56/100  =  4.8 

MgO   =  5.8X40/84  =  2.8 

ZnO            •  =6.1 


Weight  of  slag  =  75.2 

Copper  in  slag  as  intermingled  matte  =  0. 15 

Intermingled  matte  lost  in  slag  =  0.37 

Free  matte  obtained  =  5.0-0.37  =  4.63  (1) 

(2)  0  15 

"•  „    =  0.075  =  7.5  per  cent.  (2) 

z.uu 

(3)  The  calorific  power  of  the  fuel  may  be  calculated  from 
its   proximate   analysis  by    the    method    of    Goutal    (Electro- 
chemical  and    Metallurgical    Industry,   April,   1907,  p.    145),  as 
follows : 

Pure  fuel  1.0000 -0.1220  =     0.8780 

0  3290 
Per  cent  of  this  volatile     '  =     37.5  per  cent 

U.o/oU 

Calorific  power  of  volatile  matter  =  8,650  Cal. 

Calorific  power  (to  liquid  H2O) : 

Carbon         0.5490x8,100  =  4,447  " 

Vol.  matter  0.3290X8,650  =  2,846  " 


Sum    =     7,293 


504  METALLURGICAL  CALCULATIONS. 

Water  formed  =  0.45 

Latent  heat  of  vaporization  of  water 

=  0.45X606.5  =         273  Cal. 


Metallurgical  calorific  power  =     7,020    " 

Assuming  the  volatile  matter  to  be  15  per  cent  hydrogen,  40 
per  cent  oxygen  r/.ic".  45  per  cent  carbon,  the  coal  contains: 

Hydrogen  0.329X0.15  =  0.049 

Volatile  carbon  0.329X0.45  =  0.148 

Fixed  carbon  =  0.549 

Total  carbon  =  0.697 

and  the  air  necessary  for  combustion  and  products  therefrom 
are  per  unit  weight  of  fuel  used : 

Oxygen  for  C=  0.697X8/3  =      1.859  pounds. 

Oxygen  for  H  =  0.049  X  8  =     0.392      " 

Sum  =     2.251      " 
Oxygen  in  coal  =  0.329X0.40  =     0.132      " 


Oxygen  needed  from  air  =     2.119      " 

2.119X13/3X16 
Air  needed  = — ^ =      113.7  cu.  ft. 

L.Zuo 

Nitrogen  therein  =  113.7X0.792  90.0      " 

10  per  cent  surplus  air  11.4      " 

Volume  of  CO*     -     2'5;*Xl6      •         -       20.7     « 
i.yo 

,T  ,  fTT2r.  0.455X16 

Volume  of  H2O    =    — ^-^- —  =          9.0 

U.ol 

Assuming  the  tons  mentioned  are  of  2,000  pounds  each,  the 
heat  generated  per  27,000  pounds  of  fuel  used  per  day  is 

27,000X7,020  =  189,540,000  pound-Calories, 
and,  therefore,  per  100  pounds  of  ore  smelted: 

189,540,000  4-  1,000  =  189,540  pound-Calories. 

The  fuel  used  per  100  pounds  of  ore  being  54  pounds,  the 
products  of  combustion  are  54  times  the  above  calculated  vol- 
umes, to  which  must  be  added 


THE  METALLURGY  OF  COPPER.  505 

7.4  pounds  SO2  =  (7.4 X  16)  ^2.88  =  41  cubic  feet. 
6.0  pounds  SO3  =  (6.0X16) -s- 3.60  =  27  *  "       " 
6.7  pounds  SO2  =  (6.7 X  16)  *  1.98  =  54     "       " 

found  later  to  be  driven  off  the  charge.     The  stack  gases  and 
the  heat  carried  out  by  them  are  as  follows : 


CO2  613X0.59X1,000  =  361.670  ounce-cal. 

H2O  243X0.49X1,000=  119,070 

N2 

Air 

502  41X0.66X1,000=        27,060 

503  27X0.58X1,000=        15,660 


P     [  2,738X0.33X1,000  =      903,540 
ir   ) 


1,427,000 
=        89,200  pound-Cal. 

Proportion  of  the  calorific  power  of  the  fuel  in  the  hot  gases : 
89,200, 


189,540 


0.47  =  47  per  cent.  (3) 


To  find  the  heat  in  the  slag  tapped,  taking  Landis's  deter- 
mination of  302  Calories  in  slag  melted  at  its  melting  point, 
1,114°,  and  assuming  a  specific  heat  in  the  melted  state  of  0.27, 
we  have  the  heat  in  it  per  unit  weight  at  1,200°: 

302 +(0.27X86)  =  325  Cal. 
Heat  in  total  slag: 

325X75.2  =  24,440  Cal. 
Proportion  of  calorific  power  of  fuel  in  hot  slag: 

24,440 


189,540 


0.129  =  12.9  per  cent.  (3) 


To  find  the  heat  in  the  matte,  we  may  take  Landis's  deter- 
mination of  204  Calories  in  matte  just  melted  at  1,000°,  and 
assuming  a  specific  heat  of  0.14  we  have  heat  in  unit  weight  at 
1,200°: 

204 +(0.14X200)  =  232  Cal., 

and,  therefore,  heat  in  the  matte  formed: 
4.65X232  =  1,080  Cal. 


506  METALLURGICAL  CALCULATIONS. 

Proportion  of  calorific  power  of  fuel  in  melted  matte: 
1,080 


189,540 


=  0.006  =  0.6  per  cent.  (3) 


(4)  The  heat  lost  by  radiation  and  conduction  equals  all  the 
heat  brought  in  and  generated  by  combustion  and  other  chem- 
ical reactions  in  the  furnace,  minus  that  absorbed  by  chemical 
reactions  in  the  furnace,  and  minus  that  issuing  as  sensible 
heat  in  the  stack  gases,  slag  and  matte.  The  ore  mixture 
being  charged  hot,  at  350°  C.,  and  having  an  assumed  specific 
heat  of  0.15,  its  sensible  heat  is 

Calories. 

100X0.15X350  =       5,250 

Add  heat  of  combustion  of  coal         =  189,540 

Heat  available  =  194,790 

The  heats  of  chemical  reaction  of  sulphates  and  oxides  on 
sulphides  are  quite  complex,  and  we  will  show  the  calculation 
in  detail  at  the  end  of  the  problem.  Suffice  here  to  give  the  end 
results  of  the  calculation: 

Calories. 

Chemical  reactions,  net  absorption  16,480 

Heat  of  formation  of  slag,  evolved  +   9,430 

Net  heat  absorbed  in  reactions  and  combinations  7,050 

The  heat  balance  sheet  will  therefore  show: 

Available. 

Calories. 

Heat  in  hot  charges 5,250 

Heat  of  combustion  of  coal 189,540 


194,790 
Distribution. 

In  chimney  gases 89,200 

In  liquid  slag,  tapped  out 24,440 

In  liquid  matte,  tapped  out 1,080 

Absorbed  in  reactions  and  combinations 7,050 

Loss  by  radiation  and  conduction 77,020 

194,790 


THE  METALLURGY  OF  -COPPER.  507 

The  loss  by  radiation  and  conduction  per  day  will  be: 

Calories. 

77 ,020  X     270  =  20,795,000 

per  minute  20,795,000-^-1,440  14,400 

The  whole  outside  area  of  the  furnace,  including  the  base,  is 
2  (20X40) +2  (6X40) +2  (6X20)  =  2,320  sq.  ft. 

Loss  in  pound-Calories  per  square  foot  of  surface  per  minute : 

14,400^2,320  =  6.2  pound-Cal.  (4) 

(5)  The  volume  of  stack  gases,  at  0°  C.,  has  been  found  to 
be  3,662  cubic  feet  per   100  pounds  of  fuel  used,  or  3,662X 
270  =  988,740  cubic  feet  per  day  =  11.4  cubic  feet  per  second. 
At  1,000°  this  volume  would  be 

11.4X  (1,000  +  273)  4-273  =  53.2  cubic  feet. 

And  since  the  area  of  stack  cross-section  is  16  square  feet  the 
velocity  of  the  hot  gases  entering  the  stack  is 

53.2-i- 16  =  3.3  feet  per  second.  (5) 

This  is  a  low  velocity  and  shows  good  design,  which  will 
result  in  better  economy  of  fuel  than  if  high  velocity  were  used. 

(6)  The  gases  were  found  to  contain  89,200  pound-Gal,    per 
100  of  ore  smelted,  or  per  day: 

89,200X1,000  =  89,200,000  pound-Cal. 
per  hour  =     3,717,000       " 

Since  1  hp.  equals  1,400  pound-Cal.  per  hour,  we  have: 
Horse-power  at  100  per  cent  efficiency: 

3,717,000-*- 1,400  =  2,650 
Horse-power  at  7.5  per  cent  efficiency: 

2,650X0.075  =      199  hp.  (6) 

In  connection  with  requirement  (3)  of  above  problem  it  will 
be  interesting  and  instructive  to  discuss  the  chemical  and  es- 
pecially the  thermochemical  phenomena  accompanying  the 
smelting  down  of  the  ore  mixture.  The  discussion  will  be 
clearest  if  we  take  the  actual  figures  of  the  problem  in  question. 

Aside  from  the  SiO2,  A12O3,  etc.,  the  ore  mixture  contains: 


508  METALLURGICAL  CALCULATIONS. 

Per  Cent. 

Fe 10.8 

Cu 2.0 

S 5.1 

O 6.4 

and  these  four  ingredients  are  present  either  as  FeS,  FeO, 
Fe2O3,  FeSO4,  Cu2S,  Cu2O,  CuO  or  CuSO4.  With  the  above 
quantities  of  the  four  elements  in  question,  however,  the  man- 
ner in  which  they  are  combined  is  fixed  within  rather  narrow 
limits.  Each  of  the  four  binds  each  of  the  others,  and  it  can 
be  found,  by  some  patience  and  trying  out,  that  the  four  ele- 
ments, constituting  24.3  per  cent  of  the  ore  mixture,  must  be 
combined  about  as  follows  in  order  to  be  present  in  the  quan- 
tities given: 

Per  Cent 

CuO 1.0 

CuSO4 3.0 

FeS 7.3 

FeSO4 8.9     . 

FeO 0.4 

Fe2O3 3.7 

The  best  proof  of  this  statement  is  to  resolve  the  weights  of 
these  compounds  into  their  components,  and  to  thus  prove 
that  they  agree  with  the  premises. 

On  melting  this  down  to  40  per  cent  matte  we  produce 

Cu2S 2.5  per  cent  in  matte, 

FeS 2.5  per  cent  in  matte, 

SO2 7.4  per  cent  in  gases, 

FeO 11.8  per  cent  to  slag. 

The  heat  represented  by  the  formation  of  the  products  must 
be  subtracted  from  the  heat  of  formation  of  the  materials 
reacting  to  get  the  net  heat  of  the  reaction.  We  will  multiply 
the  amount  of  each  material  by  the  heat  of  formation  of  unit 
weight  from  its  elements  as  follows: 

Cat. 

CuO       1.0X(  37,700^   79.6)  =  l.OX    473  =       473 

CuSO4    3.0X  (181,700-5- 159.6)  =  3.0x1,139  =    3,417 

FeS        7.3X(  24,000^   88    )  =  7.3X    273  =     1,993 


THE  METALLURGY  OF  COPPER.  509 

FeSO  8.9 X  (214,500-152  )  =  8.9X1,411 
FeO  0.4X(  65,700-  72  )  =  0.4X  913 
Fe2O3  3.7  X  (195,600  -160  )  =  3.7X1,223 

Sum=  23,331 
Heat  of  formation  of  products: 

Cu2S  2.5  X(  20,300-159)  =    2.5  X     128  =  320 

FeS  2.5  X(  24,000-   88)  =    2.5  X    273  =  683 

SO2  7.4  X(  69,260-   64)  =     7.4X1,082  =  8,007 

FeO  11.8X(  65,700-   72)  =  11.8X    913  =  10,773 


Sum  =  19,782 
Difference,  heat  absorbed  =    3,549 

We  have,  therefore,  a  deficit,  or  heat  to  be  supplied.  This 
deficit  is  increased  by  the  heat  required  to  decompose  BaSO4 
into  BaO  and  SO3,  CaCO3  into  CaO  and  CO2  and  MgCO3  into 
MgO  and  CO2 ;  while  it  is  decreased  by  the  heat  of  combination 
of  Cu2S  and  FeS  to  form  matte,  and  the  heat  of  combination 
of  SiO2  with  BaO,  APO3,  CaO,  MgO,  ZnO  and  FeO  to  form 
slag. 

The  driving  off  of  SO3  and  CO2  absorbs 

Calories. 

SO3  from  BaO,  SO3   6.0X1,189  =    7,134 

CO2  from  CaO,  CO2    3.7X1,026  =    3,796 

CO2  from  MgO,  CO2  3.0X    666  =     1,998 


Sum  =  12,928 

The  net  result  of  all  these  reactions,  leaving  out  the  forma- 
tion heat  of  the  slag  from  its  oxide  constituents,  is 

Lb.  Cat. 

Absorbed  23,331  +  12,928  =  36,259 
Evolved  19,782 

Deficit  16,477 

And  even  if  we  credit  the  heat  of  formation  of  the  slag: 

Lb.  Cal. 
70.9X133  =    9,430 


there  remains  a  net  deficit  of      7,047 


510  METALLURGICAL  CALCULATIONS. 

The  reaction  of  the  ore  mixture  to  form  matte  and  slag  is 
therefore  an  endothermic  operation,  in  spite  of  the  fact  that 
much  sulphur  goes  off  as  SO2.  The  prime  reason  for  this  is 
that  the  bulk  of  the  sulphur  in  the  roasted  ore  was  present  as 
sulphate  and  not  as  sulphide. 

Problem  108. 

A  copper  blast  furnace  has  at  its  disposal  materials  of  the 
following  compositions  in  percentages: 

Cu.  Fe.  S.'  SiO\ 

Selected  raw  ore 15  20  35  .      25 

Roasted  concentrates 25  35  10  12 

Refinery  slag 50  10  —  25 


Limestone CaO  -  50  1 

It  is  desired  to  make  a  matte  with  50  per  cent  copper  and 
a  slag  containing  35  per  cent  SiO2,  40  per  cent  FeO  and  15 
per  cent  CaO,  or  with  these  ingredients  in  those  proportions. 

Required: 

(1)  The  proportions  of  charge. 

(2)  A  balance  sheet  showing  distribution  of  these  materials. 

Solution:  The  foregoing  conditions  are  those  which  fre- 
quently confront  the  copper  smelter.  He  has  at  hand  raw  ore 
and  roasted  concentrates,  whose  relative  quantities  he  can 
usually  vary  at  will  by  simply  concentrating  and  roasting  more 
or  less  material.  He  has  return  slags  from  the  refinery  fur- 
naces which,  however,  are  not  unlimited  in  quantity,  but  bear  a 
general  relation  to  the  weight  of  matte  made  and  sent  on  to 
the  further  operations.  Finally,  the  limestone  can  be  varied  at 
will.  It  will  readily  be  seen  that  if  the  charge  is  fixed  at  a 
certain  quantity  of  raw  ore  to  start  with,  that  there  are  then 
three  variables,  the  quantities  of  the  other  three  materials,  and 
to  fix  these  there  are  practically  only  two  conditions  to  be 
fulfilled,  the  two  ratios  between  three  components  of  the  slag. 
In  order  to  make  a  solution  possible,  it  is  necessary  to  make 
an  assumption  which  will  practically  reduce  the  number  of 
variables  by  one,  and  on  looking  over  the  ground  it  is  seen 
that  the  most  rational  assumption  which  can  be  made  is  to 
assume  the  refinery  slag  to  bear  a  given  relation  to  the  weight 
of  matte  produced.  Such  an  assumption  eliminates  the  weight 


THE  METALLURGY  OF  COPPER.  511 

of  refinery  slag  as  a  variable,  and  leaves  us  with  only  two  vari- 
ables and  two  conditions  to  fill,  which  makes  a  solution  pos- 
sible. Assuming  the  charges  to  be  based  on  100  of  roasted 
ore,  we  can  call  the  weight  of  roasted  concentrates  used  X,  the 
weight  of  refinery  slag  Y,  and  the  weight  of  limestone  Z,  and 
then  figure  out  the  weight  of  matte  produced  in  terms  of  X, 
Y  and  Z.  Assuming  then  that  the  refinery  slag  is,  say,  0.25 
of  the  weight  of  matte,  we  have  Y  =  J  (expression  for  weight 
of  matte),  and  thus  one  equation  between  X,  Y  and  Z.  The 
ingredients  of  the  slag  being  figured  out  in  terms  of  X,  Y  and 
Z,  the  assumed  relations  between  FeO,  CaO  and  SiO2  in  the 
slag  give  us  two  more  equations  between  X,  Y  and  Z,  and  thus 
all  three  quantities  can  be  determined. 

The  provisional  balance  sheet,  based  on  100  of  ore  and  X, 
Y  and  Z  of  other  ingredients  of  the  charge,  will  be  as  follows: 

BALANCE  SHEET. 

Ore.     (100).  Matte.  Slag. 

Cu         15  Cu  15 

Fe         20  Fe    7.2  +  0.14Y  FeO  16.5  -  C.18Y 

S  35  S      7.8  +  0.03X  +  0.26Y 

SiO2      25  SiO2  25.0 

R'sfd 

Cone.     (X). 

Cu      0.25X        Cu  0.25X 

Fe       0.35X        Fe0.12X  FeO  0.30X 

S         0.10X        S    0.10X 

SiO2    0.12X  SiO20.12X 

Ref. 

Slag.     (Y). 

Cu      0.50Y        Cu  0.50Y 
Fe       0.10Y        Fe  0.10Y 

SiO2    0.25Y  SiO20.25Y 

Lime- 
stone.      (Z). 

CaO    0.50Z  CaO  0.50Z 

SiO2    0.01Z  SiO20.01Z 

Notes  on  above  balance  sheet: 

Copper  in  the  matte 15     +  0.25X  +  0.50Y 

Weight  of  matte  (50  per  cent  Cu) 30     +0.50X  +  1.00Y 


512  METALLURGICAL  CALCULATIONS. 

Weight  of  S  in  matte 7.8  +  0.13X  +  0.26Y 

Weight  of  Fe  in  matte 7.2  +  0.12X  +  0.24Y 

These  weights  of  S  and  Fe  are  therefore  provided  for  under 
the  column  "  matte,"  taking  them  from  the  various  materials 
charged. 

The  refinery  slag  being  assumed  0.25  of  the  matte,  we  have 
at  once 

Y  =  0.25  (30  +  0.50X  + l.OOY) 

=  7.5  +  0.13X  +  0.25Y 
whence  Y=  10  +  0.17X 

That  is,  the  refinery  slag  equals  in  weight  0.1  the  ore  plus  0.17 
the  roasted  concentrates.  This  practically  amounts  to  leaving 
the  roasted  concentrates  and  limestone  as  the  only  variables. 

The  problem  can  now  be  solved  by  summing  up  the  ingredi- 
ents of  the  slag  as  follows: 

FeO    =  16.5-0.18Y  +  0.30X 

SiO2    =  25.0  +  0.25Y  +  0.12X  +  0.01Z 

CaO    =  0.5Z 

or  substituting   Y  =10    +0.17X 

FeO  =  14.7  +  0.27X 

SiO2  =  27.5  +  0.16X  +  0.01Z 

CaO  =    0.5Z 

And  since  the  requirements  of  the  slag  are  that 


and 

15 


FeO 


we  have: 


OK 

27.5  +  0.16X  +  0.01Z  =  ~  (14.V  +  0.27X) 


0.50Z  =        (1 

whence 

X  =  195     Z  =  48 

and,  therefore,  Y  =  43 


THE  METALLURGY  OF  COPPER.  513 

The  final  balance  sheet  then  becomes 

BALANCE  SHEET. 

Ore.  (100).               Matte.                          Slag. 

Cu  15                     Cu  15 

Fe  20                     Fe  20 

S  35                     S    35 

SiO2  25                                                    SiO2  25 

Rs't'dConc.  (195). 

Cu  49                     Cu  49 

Fe  68                     Fe  17                       FeO  66 

S  20                     S     10 

SiO2  23                                                    SiO2  23 

Ref.Slag.  (43). 

Cu  22                     Cu  22 

Fe  4                    Fe    4 

SiO2  11                                                   SiO2 11 

Limestone.  (48). 

CaO  24                                                   CaO  24 

SiO2  1                                                   SiO2    1 

172  150 

Matte.                 Per  Cent.  Slag.                          Ratio. 

Cu 86  =-50  SiO2. 60  =  36 

Fe 41  =  24  FeO 66  =  40 

S 45  =  26  CaO 24  =  14 

172  150     90 

Furnace  managers  are  usually  afraid  of  X,  Y  and  Z,  and  in 
regular  running  there  is  usually  little  need  for  an  algebraic 
solution,  yet  many  occasions  arise  when  a  judicious  use  of 
algebra  solves  a  problem  in  an  exact  manner  which  hardly 
any  amount  of  guessing  or  approximating  can  attain  efficiently. 
In  bringing  forward  this  solution  we  may  lay  ourselves  open  to 
being  called  pedantic  or  academic,  but  the  fact  is  that  the 
conditions  of  this  problem  were  proposed  to  the  writer  as  a 
difficult  nut  to  crack  by  a  practical  copper  smelter,  and  the 
algebraic  solution  furnished  was  characterized  by  him  as  the 
most  satisfactory  solution  of  this  class  of  problems  which  he 
had  yet  seen. 


514  METALLURGICAL  CALCULATIONS. 

"  Bringing  Forward  "  of  Copper  Matte. 

The  above  term  is  the  old  Welsh  expression  for  the  further 
treatment  of  mattes  which  gradually  eliminates  iron  and  sul- 
phur and  finally  results  in  "  crude  "or  "  blister  "  copper,  usually 
90  to  99  per  cent  pure.  The  matte  obtained  by  the  first  smelt- 
ing operation  varies  considerably  in  richness,  between  20  and 
50,  or  even  up  to  60  per  cent,  of  copper.  The  reason  it  is  not 
always  made  high  in  copper  is  that  the  richer  the  matte  the 
more  copper  goes  into  the  slag;  one  of  the  expert  metallurgist's 
best  accomplishments  in  copper  smelting  is  to  make  a  rich 
matte  and  poor  slag.  The  making  of  slag  of  the  best  compo- 
sition, and  the  use  of  extra  large  settlers  or  fore-hearths,  helps 
most  to  this  end.  Having  then  this  first  "  matte  "  the  problem 
of  the  metallurgist  is  to  get  from  it  the  metallic  copper  which 
it  contains. 

THE  WELSH  PROCESS. 

The  principle  employed  is  to  partially  roast  the  matte*  and 
then  to  smelt  it  down  to  a  richer  matte  and  a  ferruginous  slag. 
The  grade  of  matte  formed  in  this  smelting  depends  entirely 
upon  the  amount  of  roasting  to  which  the  matte  has  been  sub- 
jected, the  more  sulphur  eliminated  by  roasting  the  less  can  be 
present  to  form  matte  and  the  richer  the  matte. 

Illustration:  A  matte  containing  2J.. 36  per  cent  of  copper 
and  22.95  per  cent  of  sulphur  is  roasted  until  its  sulphur  con- 
tents is  two-thirds  eliminated.  What  grade  of  matte  can  be 
expected  on  the  subsequent  smelting. 

The  sulphur  left  is  22.95  -f-  3  =  7.65.  The  21.36  of  copper 
requires  5.34  of  sulphur  to  form  Cu2S,  leaving  7.65—5.34  = 
2.31  of  sulphur  to  form  FeS.  This  forms  2.31X88/32  =  6.35 
FeS.  The  resulting  matte  cannot  contain  more  FeS  than  this; 
its  composition  will  therefore  be  approximately: 

Copper 21.36  =  64  per  cent 

Iron 4.04  =  12 

Sulphur 7.65  =  23 


33.05 

In  practical  work  a  richer  matte  than  this  will  be  formed 
in  reverberatory  furnace  smelting  and  a  slightly  poorer  matte 
by  shaft  furnace  smelting,  because  in  the  former  there  is  some 


THE  METALLURGY  OF  COPPER.  515 

reaction  between  the  oxides  and  sulphides  of  the  roasted  ore, 
resulting  in  an  expulsion  of  SO2  during  smelting,  and  in  the 
latter,  using  carbonaceous  fuel,  the  temperature  is  so  high  and 
reducing  action  so  strong  that  metallic  iron  is  formed  and 
enters  the  matte,  practically  acting  as  if  the  matte  contained 
some  Fe2S,  and  thus  diluting  the  matte. 

This  roasting  is  a  slow  operation  unless  the  matte  is  crushed 
and  roasted  fine.  Lump  matte  is  very  little  pervious  to  gases; 
it  roasts  slowly.  By  breaking  a  lump  in  half  we,  on  an  aver- 
age, increase  its  surface  50  per  cent,  and  therefore  the  exposed 
surface  available  for  oxidation  increases,  for  a  given  weight 
of  material,  very  quickly  as  its  size  is  decreased.  For  auto- 
roasting  by  its  own  self-generated  heat  of  oxidation  the  finer 
the  better,  and  roasting  thus  in  lump  form  is  impracticable 
because  of  the  slowness  of  the  operation. 

Pyritic  smelting  of  matte  is  probably  altogether  out  of  ques- 
tion as  far  as  pure  pyritic  smelting  without  carbonaceous  fuel 
is  concerned.  Using  some  coke,  however,  an  oxidizing  smelt- 
ing analogous  to  partial  pyritic  smelting  is  possible,  as  was 
proved  by  Mr.  Freeland,  at  Isabella,  Tenn.  Such  a  concen- 
trating pyritic  smelting  is  more  feasible  with  a  low-grade  matte 
than  with  a  high  grade,  because  there  is  more  iron  and  sul- 
phur to  oxidize.  It  will  be  profitable  to  calculate  closely  the 
details  of  this  matte  smelting  to  rich  matte,  and  to  compare  it 
with  the  details  of  the  smelting  of  raw  ore  in  the  same  fur- 
nace— the  subject  of  Problem  105,  page  463.  ' 

Problem   109. 

W.  H.  Freeland  at  Isabella,  Tenn.  (see  Engineering  and  Min- 
ing Journal,  May  2,  1903),  smelted  a  low-grade  matte  without 
preliminary  roasting  in  a  water-jacketed  HerreshofT  furnace, 
having  a  free  area  at  the  tuyeres  of  21.7  square  feet.  The 
analyses  of  materials  used  and  the  products  are  as  follows: 

Charges. 

Raw  Laboratory 

Matte.         Ore.  Samplings.  Slags.  Quartz.  Coke. 

Cu '.     20.00         2.79  2.45         0.73         

Fe 47.15       43.26  31.07       39.20  1.45     2.30 

S 24.00       29.18  14.84         1.75  0.32     1.58 

SiO2..                   0.44       10.01  22.66       30.90  96.79     8.41 


516  METALLURGICAL  CALCULATIONS. 

Matte.         Ore.    Samplings.  Slags.    Quartz.   Coke. 
CaO..  0.10         6.32         5.71         8.51         0.23 


MgO 

1  39 

2  03 

2  71 

Zn 

2  05 

2.56 

2  05 

2  88 

APO3. 

0  82 

1  00 

1   15 

1  90 

Mn  

0.53 

0.69 

0.75 

0  85 

o   

4.91 

3.39 

11.37 

c  

13.90 

CO2,  etc  
Loss.  . 

2.80 

....  0.00 
0.32  3.56 
.  ...-  0.00 
0.38  1.00 
83.86 

0.39     .... 

Products. 

Matte.  Flue  Dust.  Slag. 

Cu 49.63  2.49  0.60 

Fe 25.24  24.79  43.99 

S 23.00  8.91  1.19 

SiO2 0.26  31.43  33.72 

CaO 3.31  2.03 

MgO 1.18  0.57 

Zn 1.53  3.81  2.12 

ArO3 3.93  2.16 

Mn..: 0.39  0.30  0.50 

0 3.97  12.86 

C 15.88 

The   charges   and  products  per  24  hours    and   per  1,000  of 
matte  used  were: 

Charges. 

Matte 47.5  tons  1,000  Ibs. 

Raw  Ore 8.1     "  170    " 

Laboratory  samplings 1.6     "  34    " 

Slag 7.6     "  160    " 

Quartz 15.7     "  330    " 

Coke 4.5     "  95    " 

Products : 

Matte 19.1  tons  401.6  Ibs. 

Flue  dust 0.6     "       .       12.0    " 

Slag 56.1     "  1182.2    " 

Blast  applied,   4,500  cubic  feet  displacement,   at   17  ounces 
pressure.     Assume  temperature  of  gases  450°  C.,  and  that  they 


THE  METALLURGY  OF  COPPER.  517 

contain  no  CO,  SO3  or  free  O2  (no  analyses  are  given).  As- 
sume matte  and  slag  issuing  from  the  furnace  at  1,300°  C.  (no 
temperature  is  given). 

Required : 

(1)  A  balance  sheet  of  everything  entering  and  leaving  the 
furnace. 

(2)  The  volume  efficiency  of  the  blowing  plant. 

(3)  The  heat  generated  per  minute  per  square  foot  of  cross- 
section  in  the  focus  of  the  furnace. 

(4)  The  theoretical  temperature  at  the  focus. 

(5)  The  proportion  of  the  heat  generated  in  the  focus  by  the 
combustion  of  carbon  and  by  the  oxidation  of  sulphides. 

(6)  The  concentration  effected  in  this  smelting. 

(1)  Balance  Sheet  (per  1000  of  matte  smelted). 

Charges.  Matte.     Flue  Dust.      Slag.  Gases. 

Matte  (1,000) 


147.82 


Cu 

200.00 

199.31 

0.30            0.39 

Fe 

471.50 

101.36 

2.97        367.17 

S 

240.00 

92.18 

....             .... 

SiO2 

4.40 

1.04 

3.36 

CaO 

1.00 



1.00 

Zn 

20.50 

6.14 

14.36 

APO3 

8.20 

.... 

8.20 

Mn 

5.30 

1.57 

3.73 

O 

49.10 

•  •  .  . 

49.10 

Ore  (170) 

Cu  4.74  4.74             

Fe  73.54  73.54 

S  49.60  ....         1.07  14.15           34.38 

SiO2  17.02  3.77  13.25             .... 

CaO  10.74  0.40  10.34             

MgO  2.36  0.14  2.22 

Zn  4.35  0.46  3.89             

APO3  1.70  0.47  1.23 

Mn  1.17  ....         0.04  1.13 

O  2.40  2.40 

CO2  2.38  ....          2.38 


518  METALLURGICAL  CALCULATIONS. 

Charges.  Matte.     Flue  Dust.  Slag.  Gases. 
Samplings  (34) 

Cu  0.83  0.83             

Fe  10.56  10.56 

S  5.05  5.05 

SiO2  7.70  7.70 

CaO  1.94  1.94             

MgO  0.69  0.69 

Zn  0.70  0.70 

A12O3  0.39  ....          ....  0.39             

Mn  0.26  .. 0.26             

O  1.15             0.48  0.67             

C  4.73             1.91             2.82 

Slags  (160) 

Cu  1.17  1.17  .... 

Fe  62.72  62.72             

S  2.80             ....  2.80 

SiO2  49.44              49.44  .... 

CaO  13.62             13.62             

MgO  4.34             4.34             

Zn  4.61  ....          4.61              

A12O3  3.04  3.04             

Mn  1 . 36  1 . 36             

O  16.90  16.90             

Quartz  (330) 

Fe  4.78  4.78             

S  1.06  1.06 

SiO2  319.41  ....          ....  319.41 

CaO  0.76  0.76             

A12O3  1.06  1.06             

H2O  2.93  2.93 


2.19  

1.50 

7.99  

3.38 

79.67 

0.27 


Coke  (95) 

Fe 

2.19 

S 

1.50 

SiO2 

7.99 

A1203 

3.38 

C 

79.67 

H2O 

0.27 

THE  METALLURGY  OF  COPPER.  519 

Charges.                           Matte.     Flue  Dust.      Slag.  Gases. 
Blast  (2,065) 

O2                 476.63             81.24  395.39 

N2               1588.77  1588.77 


(3854)  401.60       12.01       1176.05       2264.84 

Notes  on  the  Balance  Sheet. 

The  sulphur  and  carbon  of  the  charge  passing  into  the  gases 
are,  from  the  balance  sheet,  192.61  and  82.49,  respectively.  We 
can  assume  all  the  carbon  in  the  gases  as  CO2  and  all  the  sul- 
phur as  SO2  except,  say,  half  of  that  from  the  ore.  This  leaves 
192.61  —  17.19  =  175.42  of  sulphur  to  be  oxidized  at  the  focus: 

32 
Oxygen  for  sulphur  =  175  .  42  X  ™  =    175  .  42 


Oxygen  for  carbon  =   82.49X        =    219.97 


Total  to  burn  sulphur  and  carbon  at  focus  395  .  39 


Oxygen  for  Fe  in  slag  =499.31  X        =    142.66 


1  fi 

Oxygen  for  Mn  in  slag  =     6  .  48  X  -£=  =        1  .  88 


1  ft 

Oxygen  for  Zn  in  slag  =   23.56X     =  =       5.77 


Total  in  slag  150.31 

Total  oxygen  in  gas  and  slag  =   545  .  70 

Oxygen  furnished  by  solid  charges  69  .  07 

Oxygen  furnished  by  the  blast  476  .  63 

Nitrogen  furnished  by  the  blast  1588.77 

(2)  The  furnace  receives  2,065  pounds  of  blast  per  1,000  of 
matte  concentrated.     This  represents  at  0°: 

2065  X  16 

=        25,550  cubic  feet. 


520  METALLURGICAL  CALCULATIONS. 

Per  2000  Ibs.  matte         =        51,100  " 

Per  47.5  tons  matte        =  2,427,250  "  "  per  day. 

=      101,135  "  "      "    hour. 

1,686  "  "      "    minute. 

At  50°  C.  1,958  "  "      "          " 

Blower  displacement      =          4,500  "  "      "          " 


Efficiency  of  blower=  =  0.435  =  43.5  per  cent.  (2) 


(3)  To  calculate  the  heat  generated  at  the  focus  we  will 
assume  that  all  the  fixed  carbon  of  the  coke  is  there  burned 
to  CO2,  and  that  the  rest  of  the  oxygen  blown  in  produces  the 
reaction  characteristic  of  pure  pyritic  smelting.  There  is 
treated  per  minute: 

47.5X2000 

=  66  Ibs.  of  matte, 


and  the  carbon  burned  per  1000  of  matte  is  82.49  Ibs.,  generating 
82.49  X  8100  =  668,169  Ib.-Calories, 

and   absorbing   219.97  Ibs.    of  oxygen.     This   leaves   476.63  — 
219.97  =  256.67  Ibs.  of  oxygen  to  oxidize  sulphides. 

Since  by  the  "  pyritic  smelting  "  reaction,  2O2  generates  161,560 
calories,  there  is  generated  per  pound  of  oxygen  thus  used 

161,560^-64  =  2,524  Ib.-Calories, 
and  per  1000  Ibs.  of  matte  smelted  we  have 

2,524X256.67  =      677,820  Ib.-Calories, 

but  heat  generated  by  carbon  =      668,170    "          " 

Total  =  1,345,990    " 

Since  this  is  per  1000  Ibs.  of  matte  smelted,  we  have,  per  minute 

1,345,990^1,000X66  =  88,835  Ib.-Calories, 
and  per  square  foot  of  smelting  zone  area,  per  minute 

88,835  -r-  21.  7  =  4,094  Ib.-Calories.  (3) 

Comparing  this  with  the  732,930  pound-Cal.  generated  at  the 
focus  in  smelting  1,000  pounds  of  ore  and  the  3,575  pound-Cal. 


THE  METALLURGY  OF  COPPER.  521 

there  generated  per  square  foot  per  minute,  we  see  that  the 
matte  smelting  requires  more  heat  per  unit  of  charge,  princi- 
pally because  the  smelting  is  done  more  slowly.  Radiation 
losses  are  over  twice  as  great  per  unit  of  charge  treated  when 
concentrating  matte  than  when  ore  concentrating.  This  points 
to  the  utility  of  hard  driving  when  smelting  matte,  as  the 
direction  likely  to  yield  greatest  economy. 

(4)  Taking  as  a  basis  of  calculation  1,000  pounds  of  matte 
treated,  there  is  generated  at  the  focus  1,345,900  pound-Cal., 
there  is  used  2,065  pounds  of  blast,  and  there  arrives  at  the 
focus  all  of  the  charges  except  the  flue  dust,  CO2  and  H2O 
of  charges,  and  approximately  one-half  of  the  sulphur  con- 
tained in  the  raw  ore.  We  therefore  have  arriving  at  the  focus 
approximately : 

82.5  Ibs.  of  fixed  carbon, 
1133.0  Ibs.  of  sulphides. 
545.0  Ibs.  of  inert  slag-forming  material. 

These  come  to  the  focus  preheated  by  the  ascending  gases, 
and  assuming  them  to  be  preheated  to  1,000°,  we  can  find  the 
correction  to  be  used  for  their  sensible  heat: 

Carbon  82.5X380=     31,350  Ib.  Calories. 

Sulphides,  melted  1133.0X200=    226,000    "          " 

Slag-forming  material      545.0X174=      94,850    " 


352,200    " 
Heat  generated  at  the  Focus  1,345,990    " 


Total  in  the  hot  products,  at  Focus  1,698,190    "          « 

Letting  t  be  the  theoretical  temperature  at  the  focus,  then  we 
have: 

Heat  in  Slag  1176  [300+  (t  -  1100)0.27] 

"       "Matte  402  [200+  (t  -  1000)0.14] 

"       "  S  vapor  17  [179+ (t- 445)0.11] 

OC1 

«       «  SQ2  _J±££ 

Z .  oo 

"       "  C°2 


522  METALLURGICAL  CALCULATIONS. 

The  calorific  capacity  of  these  products  at  t°  is  therefore 

29,858  +  858.7t  +  0. 1046t2 
and  making  this  equal  to  the  heat  available  at  the  focus,  we  have 

0.1046t2  +  858.7t  +  29,858  =  1,698,190 
whence  t  =  1622°.  (4) 

This  is  53°  higher  than  we  calculated  for  the  same  furnace 
smelting  ore,  with  only  one-third  as  much  coke.  If  the  coke 
were  omitted  from  this  charge  the  theoretical  temperature 
would  be: 

t  =  1388°. 

Such  a  theoretical  temperature  at  the  focus  would  not  suffice 
to  form  slag,  supply  radiation  and  conduction  losses,  and  see 
the  matte  and  slag  safely  out  of  the  furnace,  except  with  very 
hard  driving  and  very  fast  running  in  a  large  furnace. 

(5)  The  heat  generated  by  oxidation  at  the  focus  has  already 
been  calculated.  The  proportion  to  credit  to  carbon  is 

668'17°        0.496  =  49.6  per  cent. 


1,345,990 
and  to  oxidation  of  sulphides  60.4  (5) 

In  the  ore  smelting  these  figures  were  found  to  be  31  and  69 
per  cents,  respectively. 

(6)  The  ratio  of  concentration  is  usually  found  by  com- 
paring the  per  cent  of  copper  in  the  material  treated  with  that 
in  the  material  produced.  This  would  give  in  this  case: 


(6) 


20.00 


A  more  reasonable  factor,  however,  is  the  ratio  of  weight  of 
fresh  copper-bearing  materials  treated  to  weight  of  copper- 
bearing  materials  produced  which  must  be  further  treated. 
This  would  be  in  this  case: 


THE  METALLURGY  OF  COPPER.  523 


1204  0  A 
-j-i-p  =  2.9 
414 


whereas  in  the  ore  smelting  it  was 
1080 


148 


=  7.3 


"Blister-Roasting"  or  "Roasting-Smelting." 

When  a  matte  has  been  concentrated  to  somewhere  between 
70  and  80  per  cent  of  copper.it  is  called  "  white  metal,"  and  is 
in  shape  for  reduction  to  metallic  copper: 

Cu2S.      FeS. 

With  70  per  cent  copper,  matte  contains. . .   87.5       12.5 
With  80  per  cent  copper,  matte  contains. . .  100.0         0.0 

The  problem  being  to  get  metallic  copper,  the  operation  is 
entirely  one  of  oxidation;  first,  slow  melting  down  of  the  lumps 
of  matte  in  a  highly  oxidizing  atmosphere  on  the  hearth  of  a 
reverberatory  furnace;  second,  continued  oxidation  until  all 
the  sulphur  is  removed  and  the  bath  remains  as  metallic  copper, 
saturated  with  Cu2O  and  Cu2S. 

During  the  melting  down  the  matte  is  oxidized  superficially 
to  such  an  extent  that  when  fusion  is  finally  complete  the  cop- 
per oxides  have  reacted  upon  the  iron  sulphides  sufficiently  to 
eliminate  all  the  iron  from  the  matte.  The  reactions  and  phe- 
nomena of  this  period  are  exactly  those  of  the  matte  concen- 
tration processes. 

After  the  melting  down  sulphur  continues  to  be  oxidized 
with  formation  of  copper  oxide,  and  then  this  latter  reacts  with 
more  of  the  sulphide  to  set  free  metallic  copper. 

(a)  2Cu2S  +  3O2  =  2Cu2O  +  2SO2. 

(b)  Cu2S  +  2Cu2O  =  6Cu  +  SO2. 

(c)  Cu2S  +  O2  =  Cu  +  SO2. 

Reactions  (a)  and  (b)  are  probably  consecutive,  but  assuming 
them  simultaneous  we  have  equation  (c). 


524  METALLURGICAL  CALCULATIONS. 

Thermochemically,  equation  (a)  analyzes  as  follows: 

2(Cu2,  S)       =2(20,300)  =    40,600  Calories  absorbed. 

2(Cu2, 0)      =*  2(43,800)  =    87,600        "        evolved. 

2(S,  O2)         =2(69,260)  =138,520        u 


Algebraic  sum  =  185,520 
Per  kilo,  of  Cu2S  =          580 

Similarly,  equation  (b) : 

(Cu2,  S)  =     20,300  Calories  absorbed. 

2(Cu2, 0)      =  2(43,800)  =    87,600 

(S,  O2)  =     69,260        "        evolved: 


Algebraic  sum  =  38,640  Calories  absorbed. 
Per  kilo,  of  Cu2S                         =         243 

Equation  (c)  gives: 

(Cu2,  S)  =  20,300  Calories  absorbed. 

(S,  O2)  =  69,260        "      evolved. 


Algebraic  sum  =    48,960 
Per  kilo,  of  Cu2S  308 

Per  kilo,  of  copper  liberated     =          385 


u  u 


The  net  result  of  these  figures  is  to  show  that  the  first  re- 
action evolves  a  large  amount  of  heat,  that  the  second  absorbs 
considerable,  but  that  the  two  together  constitute  a  highly  exo- 
thermic reaction.  When  we  reflect  that  a  kilogram  of  melted 
Cu2S  only  carries  some  250  Calories,  and  a  kilogram  of  melted 
copper  not  over  200  at  any  furnace  temperature,  the  excess 
of  heat  above  calculated  shows  up  very  strikingly. 

In  a  reverberatory  furnace  the  rate  of  oxidation  is  so  slow 
that  but  minor  advantage  is  taken  of  the  heat  of  oxidation  of 
the  Cu2S. 

Illustration:  In  a  reverberatory  furnace,  8  tons  of  "white 
metal  "  is  smelted  to  blister  copper  in  48  hours,  using  5  tons 
of  coal.  About  what  proportion  does  the  heat  of  oxidation  of 
the  charge  bear  to  the  heat  of  combustion  of  the  coal? 

Assuming  the  tons  to  be  1,000  kilos.,  the  white  metal  to  be 


THE  METALLURGY  OF  COPPER.  52.5 

nearly  pure  Cu2S,  and  the  calorific  power  of  the  coal  6,000,  we 
have: 

Heat  of  oxidation  of  the  bath  8000  X  308  =    2,464,000  Calories. 
"     of  combustion  of  coal       5000  X  6000  =  30,000,000        " 


"     requirement  for  48  hours  32,464,000       " 

"     requirement  per  hour  676,333       " 

Proportion  of  heat  requirement  furnished  by  oxidation  of 
bath: 

2,464,000 


34,920,000 


0.759  =  7.6  per  cent. 


The  above  figures  teach  us,  however,  that  when  the  charge  is 
once  melted,  if  the  oxidation  of  the  bath  could  be  per- 
formed quickly  enough  it  alone  would  keep  the  furnace  up  to 
heat.  For  example,  the  furnace  requires  an  average  of  727,500 
Cal.  per  hour  to  keep  it  up  to  heat.  The  heat  of  oxidation 
would  therefore  supply  this  for 

2,464,000 
727,500     =3 

which  means  that  if  the  bath,  when  melted,  could  be  oxidized 
in  that  time  all  exterior  firing  would  be  unnecessary  after  the 
charge  had  been  once  melted.  Some  companies  force  air  onto 
the  surface  of  the  bath,  and  one  company  blows  compressed 
air  through  wrought  iron  pipes  into  the  bath,  producing  the 
required  oxidation  in  one-fifth  the  time  usually  required,  and 
saving  greatly  in  coal. 

Bessemerizing  Copper  Matte. 

John  Hollaway,  in  1878,  patented  the  process  of  oxidizing 
matte  to  metallic  copper  in  an  apparatus  similar  to  the  Bes- 
semer steel  converter;  in  1880,  Manhes,  in  France,  was  suc- 
cessful in  accomplishing  this  result,  and  in  1884  ran  the  first 
commercial  plant  at  the  Parrot  works  in  Butte.  While  the 
principles  are  similar  the  details  are  very  different  from  the 
blowing  of  pig  iron  to  steel. 

In  oxidizing  pig  iron  the  carbon  silicon,  manganese,  etc., 
which  are  to  be  oxidized  out  rarely  exceed  10  per  cent,  while 


526  METALLURGICAL  CALCULATIONS, 

the  iron  itself  is  oxidizable,  and  some  3  to  20  per  cent  may  be 
lost.  In  oxidizing  matte  some  40  to  70  per  cent  of  the  whole 
charge  is  to  be  oxidized,  a  very  voluminous  slag  results,  and 
the  operation  lasts  five  to  fifteen  times  as  long.  Towards  the 
end  of  a  "  steel  "  blow,  the  pure  iron  formed  is  itself  oxidizable, 
and  is  not  chilled  but  really  heated  by  the  passing  of  the  blast 
through  it;  in  a  "  matte  "  blow  the  pure  copper  separating  at  the 
end  is  not  oxidizable  under  the  prevailing  conditions,  and 
therefore  is  only  chilled  by  the  blast  if  the  latter  strikes  it. 
On  the  latter  account,  tuyeres  in  the  bottom  are  impracticable 
when  Bessemerizing  matte,  since  they  become  filled  with  chilled 
copper;  it  is  imperative  to  use  lateral  tuyeres  which,  by  the 
swinging  of  the  converter,  are  kept  always  below  the  surface  of 
the  matte  but  above  the  pool  of  copper  as  it  separates  out  and 
sinks  to  the  bottom. 

The  two  best  treatises  for  details  of  bessemerizing  copper 
matte  are  *Jannettaz's  "Les  Convertisseurs  pour  le  Cuivre"  and 
Dr.  F.  Mayr's  "  Das  Bessemern  von  Kupf ersteinen ;  "  very  satis- 
factory information  can  be  found  in  Dr.  Peters'  "  Modern 
Copper  Smelting "  and  "  Principles  of  Copper  Smelting,"  in 
Hixon's  "  Notes  on  Lead  and  Copper  Smelting  and  Copper 
Converting,"  and  in  Schnabel's  "  Handbook  of  Metallurgy," 
last  edition,  Vol.  I. 

Assume  that  a  converter  is  just  emptied,  is  at  a  bright  red 
heat  (1,100°),  and  that  a  charge  of  melted  matte  at,  say,  1,100° 
(setting  point  1,000°),  is  poured  into  it.  The  lining  is  some 
60  cm.  thick,  the  outside  shell  of  the  converter  is  at  about 
200°  C.,  and  the  surface  is  losing  heat  at  a  nearly  steady  rate 
of,  say,  50  Cal.  per  square  meter  of  outside  surface  per  minute 
(10  pound-Cal.  per  square  foot).  A  converter  of  25  square 
meters  outside  surface  would  thus  lose  1,250  Cal.  per  minute 
if  standing  still.  If  it  had  3,000  kg.  of  liquid  matte  in  it,  with 
a  specific  heat  of  0.14,  the  latter  would  give  out  3,000x0.14  = 
420  Calories  for  every  degree  which  it  cooled,  and  would  there- 
fore cool  off  about  1, 250 -v- 420  =  3°  per  minute.  If  poured 
in  100°  above  its  setting  point  it  might  be  some  30  minutes  in 
cooling  to  its  setting  point  and  70  minutes  in  setting,  assuming 
no  radiation  losses  through  the  throat,  i.e.,  that  the  throat 
were  tightly  covered. 

*  Mem.  de  la  Soc.  Ing.  Civile,  1902,  (I),  268-319. 


THE  METALLURGY  OF  COPPER.  527 

Under  such  conditions,  if  blast  is  turned  on,  the  reaction  is 
2FeS  +  3O2  +  2SiO2  (lining)  =  2(FeO,  SiO2)+2SO2 

and  the  heat  evolved  is 

Decomposition  of       2FeS  =  2(24,000)  =  -  48,000  Calories. 

Formation  of  2FeO  =2(65,700)  =+131,400       " 

Union  of  2FeO  with  2SiO2  =  2(8,900)  =  +    17,800 

Formation  of  2S02  =2(69,260)  =+138,520 


Total  =+239,720 

Heat  evolution  per  1  kg.  FeS  1,362  Calories. 

Heat  evolution  per  1  kg.  O2  2,497        " 

THEORETICAL  TEMPERATURE  RISE. 

Supposing  we  oxidize  an  amount  of  FeS  equal  to  1  per  cent 
of  the  weight  of  matte.  Let  us  calculate  the  theoretical  rise 
in  temperature  of  the  contents  of  the  converter. 

The  oxygen  required  is  3O2  to  2FeS,  or  per  kilo,  of  FeS. 

Oxygen  required  =  96-7-176     =  0.545  kg. 
N2  accompanying  =1.818    " 

Air  required  =  2.363    " 

Volume  =  2.363-7-1.293  =  1.827m3. 

In  being  raised  from  say  50°  C.  to  1100°,  the  assumed  tempera- 
ture of  the  matte,  the  air  will  absorb 

1.827[0.303  + 0.000027(1150)]  X  1050  =  641  Calories. 

This  leaves  available,  for  raising  the  temperature  of  the  pro- 
ducts of  the  reaction: 

1,362-641  =  721  Calories. 
The  immediate  products  are: 

99.      kg.  of  unoxidized  matte. 
1.50    "     "  liquid  slag. 
0.73    "     "  SO2  gas. 
1.82    u     "  N2gas, 


528 


METALLURGICAL  CALCULATIONS. 


The  heat  capacities  of  these,  at  1100°,  are  as  follows,  per  1°  C. : 


Matte 

Slag 

SO2 

N2 


99X0.14  =  13.86  Calories. 
1.50X0.27  =    0.41 
0-73-f-2.88Xl.02  =    0.25 
1.8184-1.26X0.36  =    0.52       " 


Sum  =  15.04 


Theoretical  temperature  rise 
721-15.04 


47.9°  C. 


The  above  rise  represents  the  rate  at  which  the  temperature 
tends  to  rise  at  the  beginning  of  the  blow.  Supposing  this 
amount  of  FeS  (1  per  cent)  is  oxidized  in  1  minute,  the  tem- 
perature at  the  end  of  1  minute  would  rise  47.9°  less  the  cooling 
off  in  1  minute,  which  latter  might  be  from  2°  to  10°,  according 
to  the  size  of  the  converter  and  its  charge,  thickness  of  lining, 
etc.  For  practical  purposes  the  cooling-off  rate  may  be  as- 
sumed as  nearly  constant,  but  the  heating-up  rate  is  quite 
variable.  As  the  FeS  disappears  the  amount  of  slag  increases, 
while  that  of  the  matte  decreases;  but  since  1.50  kilos,  of  slag 
with  a  calorific  capacity  of  0.41  Cal.  per  degree  takes  the  place 
of  1  kilo,  of  matte  with  a  calorific  capacity  of  0.14  Cal.  per  1°, 
the  heat  capacity  of  the  products  is  increasing  0.27  Cal.  for 
each  1  per  cent,  of  FeS  oxidized,  while  the  available  heat  for 
raising  temperature  is  slightly  decreasing,  because  of  the  higher 
temperature  of  the  bath  and  therefore  the  greater  chilling  effect 
of  the  air.  Assuming  that  the  bath  cools  5°  during  the  time 
that  1  per  cent  of  FeS  is  being  oxidized  out,  we  can  calculate 
the  following  table: 

Calorific 


Temp. 

Heat           Net 

Capacity 

FeS 

at 

Absorbed       Heat 

"f 

Temp. 

Oxidized. 

Start. 

by  Air.  Available. 

Products. 

"Rise. 

Oto    1% 

1100° 

641           715 

15.04 

42.9° 

1   "     2% 

1142.5° 

670           686 

15.31 

40° 

2  «     3% 

1182° 

694           662 

15.58 

37° 

3^     4% 

1219° 

722           634 

15.85 

35° 

4  "     5% 

1254° 

742           614 

16.12 

33° 

5  "     6% 

1287° 

768          588 

16,39 

31° 

THE  METALLURGY  OF  COPPER.  529 

Calorific 

Temp.  Heat          Net      Capacity 

FeS                  at  Absorbed       Heat            of  Temp. 

Oxidized.          Start.  by  Air.  Available.  Products.  Rise. 

6  "     7%         1318°  789           567           16.66  29° 

7  "     8%         1347°  809           547           16.93  27° 

8  "     9%         1374°  823           533           17.20  26° 

9  "   10%         1405°  847           509           17.47  24° 
At     10%         1429°  

There  is  no  object  in  extending  above  table,  because  it  is 
constructed  on  the  particular  assumption  that  radiation  losses 
would  amount  to  5°  during  the  burning  out  of  1  per  cent  of 
FeS.  This  quantity  would  evidently  vary  with  the  size  of  the 
converter,  the  amount  of  matte  being  treated  and  the  speed  of 
blowing,  because  as  the  contents  become  hotter  their  rate  of 
cooling  would  be  increased.  The  object  of  the  above  table,  so 
far  as  it  went,  was  to  show  that  47.5°  was  the  theoretical  rise 
for  the  first  1  per  cent,  but  that  the  theoretical  rises  for  suc- 
ceeding per  cents  would  be  less  and  less;  in  such  manner,  in- 
stead of  rising  (47.9  -  5)  X 10  =  429°  for  an  oxidation  of  10 
per  cent  of  FeS  the  actual  rise  figures  out  only  329°. 

One  of  the  most  necessary  data  which  is  badly  needed  for 
these  converters,  and  in  fact  for  Bessemer  converters  in  gen- 
eral, is  the  heat  loss  by  radiation  and  conduction,  in  other 
words,  how  much  heat  would  be  lost  by  the  charge  if  simply 
standing  still,  how  much  would  the  temperature  of  a  given 
charge  fall  per  minute  if  standing  still.  This  could  be  easily 
obtained  by  running  in  a  charge  of  pretty  hot  matte,  and  fol- 
lowing its  tempera tue  curve  as  it  cools,  without  any  necessity 
of  letting  it  freeze,  but  merely  starting  up  the  blast  when  the 
rate  of  cooling  has  been  satisfactorily  determined.  If  these 
determinations  were  coupled  with  details  as  to  the  temperature 
of  the  outside  air,  its  velocity  of  impingement  against  the  con- 
verter, the  temperature  of  the  outside  shell,  the  area  of  radiating 
surface  and  the  thickness  of  the  lining,  we  would  soon  get  data 
with  which  to  render  entirely  definite  and  exact  the  whole 
thermal  investigation  of  a  "  Bessemerizing  "  operation.  This 
should  be  supplemented  by  analyses  of  the  gases  during  the 
blow  and  the  temperature  curve  of  the  contents  as  the  blow 


530 


METALLURGICAL  CALCULATIONS. 


progresses.  Scientific  metallurgists  must,  at  present,  simply 
assume  many  of  these  data,  because  of  lack  of  them.  We  are 
looking  to  the  metallurgical  directors  of  copper  plants  for  some 
of  this  badly-needed  technical  data. 

Problem  110. 

W.  Randolph  Van  Liew  (Trans.  Am.  Inst.  Mining  Eng., 
1904,  p.  418)  gives  the  following  analyses  of  a  charge  of  matte 
blown  to  blister  copper  in  a  Bessemer  converter: 


Cu. 
Matte  49  .  72 

Fe. 
23.31 

5. 
21.28 

Zn. 

1.19 

10  minutes       50.20 

23.15 

20.95 

1.20 

20      "               56.88 

17.85 

19.74 

0.84 

30      "               64.60 

10.50 

18.83 

0.70 

40      "  (last  slag  skimmed)  76  .  37 
70      "      (blister  copper)  99.120 

2.40 
0.038 

16.30 
0.159 

0.45 
0.09 

56. 

Ag. 

An. 

0.14 

0.152 

0.00055 

0.12 

0.147 

0.00048 

0.10 

0.176 

0.00069 

0.13 

0.191 

0.00083 

0.13 

0.240 

0.00110 

0.006 

0.312 

0.00111 

As. 

Matte 0.11 

10  minutes 0.09 

20      "         0.08 

30      "         0.08 

40      "  (last  slag  skimmed)  0 . 08 

70      "       (blister  copper)  0.001 


The  percentage  composition  does  not  exhibit  clearly  the 
relative  time  and  amount  of  the  elimination  of  impurities,  be- 
cause of  the  varying  weight  of  the  bath. 

Required : 

(1)  Assuming  1,000  kilograms  of  matte  to  be  treated,  find 
the  weight  of  the  matte  at  each  period  of  the  blow. 

(2)  The  loss  of  each   constituent  of  the  bath   during   each 
period. 

(3)  The  heat  evolution  during  each  period. 

(4)  The  loss  of  heat  per  minute  due  to  radiation  and  con- 
duction, assuming  that  the  heat  starts  with  matte  at  1,100°  and 
ends  with  blister  copper  at  1,200°,  and  that  1  per  cent  of  copper 
(reckoned  on  the  matte)  is  oxidized  during  the  last  period. 


THE  METALLURGY  OF  COPPER.  531 

(1)  The  first  question  is  to  find  some  constituent  of  the 
matte  whose  weight  does  not  vary  during  the  blow,  to  serve 
as  a  basis  for  the  calculations.  If  the  analyses  be  taken  as  re- 
liable in  all  details  (we  can  make  no  other  assumption)  we 
see  that  there  is  very  little  loss  of  anything  in  the  first  10  min- 
utes, evidently  because  of  the  low  temperature  of  the  matte. 
Afterwards,  iron  falls  off  rapidly,  also  sulphur  and  zinc,  while 
silver  and  gold  increase  in  percentage,  because  of  the  falling  off 
in  weight  of  the  bath  as  a  whole.  The  slag  up  to  the  last  skim- 
ming, contains  usually  but  very  little  copper;  in  the  last  period 
we  are  told  to  assume  a  loss  of  10  kilos  of  copper  by  oxidation. 
The  most  rational  basis  for  calculating  the  weight  of  the  bath 
is  to  assume  the  copper  contents  constant  for  the  first  40  minutes. 

Copper  in  1000  kg.  of  matte  at  starting  =     497.2  kg. 
Weight  of  matte  at  starting  1000.0    " 

"'     "       "       10  minutes  =  497.2-- 0.5020  =     990.4    " 

"•      20     ."         =  497.2-0.5688  =    874.1    " 

"       30        "         =  497.2-0.6460  =     769.7    " 

"       "       40        "         =497.2-0.7637  =    651.0kg. 

"   metal  70        "         =  487.2-0.9912  =     491.5    " 

(1) 

From  the  analyses  given,  and  calling  the  shortage  of  per- 
centage in  the  analyses  oxygen,  we  have  the  following  table 
of  weights  and  eliminations:  (2) 

Per  1000  kg.  of  Original  Matte. 


Start  

Cu. 
497.2 

Fe. 
233.1 

5. 
212.8 

0. 
41  0 

Eliminated 

0  0 

3.8 

5  3 

0  1 

End  10'  

497.2 

229.3 

207  5 

40  9 

Eliminated 

0  0 

73  3 

35  0 

2  9 

End  20'.... 

497  2 

156  0 

172  5 

38  0 

Eliminated  
End  30'  

0.0 

497.2 

75.2 

80.8 

27.6 

144  9 

-0.3 

38  3 

Eliminated  
End  40'  

0.0 

497.2 

65.2 

15.6 

38.8 

106  1 

12.0 

26  3 

Eliminated  
End  70'.. 

10.0 

487.2 

15.4 

0.2 

105.3 

0.8 

24.9 
1.4 

532  METALLURGICAL  CALCULATIONS. 

Zn.  As.  Sb.  Ag.  An. 

11.9  1.1  1.4  1.52  0.0055=1000.0 

0.0  0.2  0.2  0.06  0.0007  =       9.6 

11.9  0.9  1.2  1.46  0.0048=990.4 

4.6  0.2  0.3  —0.08  —0.0012  =  116.3 

7.3  0.7  0.9  1.54  0.0060=874.1 
1.9  0.1  —0.1  0.07  —0.0004=104.4 

5.4  0.6  1.0  1.47  0.0064=   769.7 

2.5  0.1  0.2  -0.09  —0.0008  =   118.7 
2.9  0.5  0.8  1.56  0.0072  =   651.0 
2.5  0.5  0.8  0.03  0.0017  =  159.5 
0.4  0.0  0.0  1.53  0.0055=491.5 

Our  table  is  not  absolutely  accurate,  as  can  be  seen  from  an 
apparent  gain  in  both  silver  and  gold  in  the  periods  10'  —  20' 
and  30'  —  40'.  This  is  caused  by  a  loss  of  copper  during  those 
periods,  but  the  amounts  of  gold  and  silver  are  too  small  to 
serve  as  a  base  for  revising  the  table.  If  we  were  to  assume 
the  gold  as  constant  it  would  make  a  smaller  weight  of  bath 
at  the  ends  of  those  periods,  but  the  figures  are  not  reliable 
enough  to  make  this  correction  worth  while. 

The  very  small  total  loss  in  the  first  10  minutes  and  the 
loss  of  time  thus  occasioned,  could  in  all  probability  be  obviated 
by  putting  the  matte  hotter  into  the  converter  at  the  start. 

A  graphic  representation  of  the  course  of  a  blow  is  usually 
made  by  plotting  the  percentages  of  each  element  in  the  bath 
at  given  periods.  This  plan  is  largely  misleading',  the  diagram 
should  be  made  by  plotting  the  calculated  weights  of  each 
element  present  at  the  given  period,  such  as  are  found  in  the 
above  table. 

(3)  The  heat  evolution  is  to  be  found  by  taking  the  weights 
of  each  element  oxidized  out,  calculating  the  heat  of  its  oxida- 
tion and  formation  of  slag,  and  subtracting  the  heat  necessary 
to  break  up  the  equivalent  amount  of  its  sulphide.  The  heat 
necessary  to  separate  the  sulphides  from  each  other  is  unknown. 

Period  /.—Start  to  10'. 

Heat  of  oxidation:  Cal.     Cat. 

Fe  to  FeO.SiO2  3.8X1332       =    5,062 

S     "  SO2  5.3X2164       =  11,469 


THE  METALLURGY  OF  COPPER.  533 

Cat.     Cal. 

As  "  As2O3  0.2X1043       =       209 

Sb  "  Sb2O3    .  0.2X695         =        139 

16,879 

Decomposition  of  sulphides: 

FefromFeS  3.8X429 

As     "     As2S3  0.2X2000(?)  = 

Sb     "     Sb2S3  0.2X1433       = 


Net  heat  evolution  14,562 

The  other  periods  are  similarly  calculated,  and  yield  the  fol- 
lowing results: 

Start-10'   10'-20'  20'-30'  30'-40'  40'-70'     -}  last. 

Heat  of  oxida- 
tion  16,879  179,797  162,476  174,316  257,046  85,682 

Decomp.  of  sul- 
phides   2,317     35,321     33,719     30,113    ,10,410     3,470 


Net  heat  evolu- 
tion  14,562  144,476  128,757  144,203  246,636  82,212 

The  last  column  is  added  for  comparison,  being  the  heat 
evolution  per  average  10'  in  the  last  period.  (3) 

(4)  The  loss  of  heat  by  radiation  and  conduction  can  be 
found,  as  a  whole,  by  assuming  the  converter  body  to  contain 
the  same  heat  at  finishing  as  at  starting,  which  is  a  likely  as- 
sumption, since  the  lining  loses  somewhat  in  weight  but  ends 
up  at  a  higher  temperature.  Then  we  know  how  much  heat 
was  in  the  original  matte  at  1,100°,  how  much  was  generated, 
how  much  is  in  the  slag  and  copper,  and  can  calculate  ap- 
proximately how  much  is  carried  out  in  the  gases.  These 
enable  us  to  find,  by  difference,  the  loss  by  radiation  and  con- 
duction which  can  then  be  averaged  up  per  minute. 

Calories. 

Heat  in  1000  of  matte  at  1100°  =  214,000 

Net  heat  generated  in  the  blow  =  678,635 

Total  available  =  892,635 


534  METALLURGICAL  CALCULATIONS. 

Heat  in  491. 5  Copper,  at  1200°  -    85,995 

"      "  424.0  SO2,        "   1000°  =    97,020 

"      "  947.    N2,          "   1000°  =  248,160 

"      "  550.    Slag,       "  1250°  =  187,000 


Heat  accounted  for  618,175 


Loss  by  radiation  and  conduction  =  274,460 

Loss  per  minute,  per  1000  kg.  matte  =      3,920    (4) 

It  is  thought  that  the  principles  explained  and  the  methods 
of  calculation  illustrated  will  suffice  to  show  to  copper  metal- 
lurgists how  important  information  is  obtainable  by  applying 
the  principles  of  thermochemistry  to  copper  smelting,  and  to 
indicate  the  lines  along  which  experiment  and  results  of  meas- 
urement and  observation  are  greatly  to  be  desired. 


THE  ELECTROMETALLURGY  OF  COPPER. 

The  electric  current  is  used  in  metallurgy  either  for  its  elec- 
trolytic effect  or  for  its  electrothermal  action.  In  the  former 
the  property  of  the  current  utilized  is  its  ability  when  pass- 
ing in  one  direction  through  an  electrolyte,  of  causing  at  the 
cathode  a  reducing  action,  such  as  the  separation  of  metal 
from  the  electrolyte  or  the  reducing  of  a  ferric  salt  to  a  ferrous 
salt,  and  at  the  anode  a  perducing  effect,  the  direct  opposite 
chemically  of  reducing,  such  as  the  taking  of  a  metal  into  the 
electrolyte  or  the  perducing  of  a  ferrous  or  cuprous  salt  to  a 
ferric  or  cupric  condition.  In  such  a  process  heat  is  inevitably 
generated  to  some  extent  by  the  passage  of  the  current  through 
the  ohmic  resistance  offered  by  the  electrolyte.  The  extent  to 
which  heat  is  thus  generated,  coincident  with  the  electrolytic 
action  of  the  current,  is  of  no  significance  whatever  upon  the 
nature  of  the  process,  which  remains  essentially  electrolytic 
as  long  as  the  current  is  used  for  and  performs  its  electrolytic 
decomposing  function.  However,  when  the  heat  thus  coin- 
cidently  and  inevitably  generated  in  the  operation  of  a  process 
essentially  electrolytic,  is  sufficient  to  keep  melted  an  electrolyte 
which  is  not  liquid  at  ordinary  temperatures,  the  apparatus  as 
a  whole  may  not  improperly  be  regarded  as  a  furnace,  and  is 
very  properly  classed  as  an  electrolytic  furnace. 


THE  METALLURGY  OF  COPPER.  535 

Electrothermal  processes  are  those  characterized  by  the 
absence  of  electrolysis,  as  shown  by  the  arrangement  and 
working  of  the  apparatus,  use  of  alternating  current,  etc.,  and 
in  which  the  current  is  used  solely  for  its  heating  effects.  Such 
processes  are  carried  on  in  electric  furnaces,  and  are  essentially 
processes  in  which  chemical  reactions  or  physical  changes  are 
brought  about  in  the  charge  solely  by  the  action  of  the  tem- 
perature maintained  by  the  assistance  of  the  electric  current. 
Metallurgically,  the  furnaces  used  are  resistance  furnaces,  arc 
furnaces  and  combinations  of  the  two.  In  resistance  furnaces 
the  heat  operating  the  furnace  is  produced  through  the  agency 
of  the  resistance  of  the  substance  or  charge  itself,  or  of  a 
solid,  liquid  or  granular  resistor,  intermingled  with,  in  contact 
with,  or  placed  in  the  neighborhood  of  the  material  to  be  heated. 
In  arc  furnaces  the  current  jumps  a  gap  between  two  poles, 
and  generates  locally  the  very  high  temperature  of  the  arc, 
which  is  utilized  by  feeding  the  material  to  be  treated  into 
it  or  by  bringing  in  close  proximity  to  it.  In  the  combined 
arc-resistance  furnace,  the  material  being  treated  forms  one  or 
both  poles  of  the  arc,  and  is  therefore  heated  by  the  arc  itself 
as  well  as  by  the  passage  of  the  current  through  its  own  substance. 

In  electrolytic  processes  the  output  of  material,  or  com- 
mercial efficiency  of  the  process,  is  essentially  dependent  upon 
the  amperage  of  the  current,  since  electrolytic  effects  are  pro- 
portional to  the  amperes  passing  through  the  electrolyte.  In 
electrothermal  processes  the  commercial  efficiency  of  output 
will  vary  with  the  total  energy  dropped  by  the  current  in  the 
furnace,  i.  e.,  will  be  proportional  to  the  watts  of  current  used 
up,  not  to  its  amperage  or  voltage,  but  to  the  product  of  these. 
Direct  currents  only  are  employed  in  electrolytic  processes; 
direct  or  alternating  may  be  used  in  electrothermal  processes, 
but  alternating  are  preferred,  because  of  the  complete  absence 
of  one-sided  electrolytic  effects  when  they  are  used. 

Assuming  familiarity  with  the  ordinary  non-electric  metal- 
lurgy of  copper,  we  may  divide  the  electrometallurgy  of  copper 
into  the  following  classes: 

I.    Electrolytic  processes — 

1.  Direct  treatment  of  ores. 

2.  Treatment  of  matte. 


536  METALLURGICAL  CALCULATIONS. 

3.  Extraction  from  solutions. 

4.  Refining  of  impure  copper. 

II.     Electrothermal  processes — 

1.  Direct  smelting  of  ores. 

2.  Melting  and  casting  of  copper. 

I. 
Electrolytic  Processes. 

Copper  has  an  atomic  weight  of  63.6.  It  occurs  chemically 
as  cuprous  compounds,  formulae  CuA1  ,  or  cupric  compounds, 
formulas  CuA11 ,  where  A1  is  a  univalent  or  monad  acid  radicle 
and  A"  a  bivalent  or  dyad  acid  radicle.  As  a  monad  atomf 
copper  has  a  chemical  equivalent  of  63.6,  as  a  dyad  clement 
31.8.  The  amounts  of  copper  dissolved  into  or  deposited  from 
a  cupric  or  cuprous  salt  are  proportional,  to  the  chemical  equi- 
valent of  copper  in  these  two  states  and  to  the  amperes  flowing. 
Assuming  that  1  ampere  liberates  electrolytically  0.00001036 
grams  of  hydrogen  per  second,  we  will  have  as  the  amount 
of  copper  concerned  in  the  passage  of  1  ampere  through  one 
tank: 

Cuprous  Compounds.     Cupric  Compounds. 
1  Ampere,  per  second . .    0.0006589  grams        0.0003295  grams 
per  minute  . .    0.03953         "  0.01977 

per  hour 2.372  "  1.186 

per  day 56.93  "  28.46 

per  year 20.78     kilograms.      10.39        kilograms 

A  useful  datum  to  remember,  if  one  is  used  to  working  in 
English  measures,  is  that  one  ampere  deposits  practically  one 
ounce  avoirdupois  (28.35  grams)  of  copper  per  day  in  each  cell 
using  cupric  compounds,  and  2  ounces  for  cuprous  compounds, 
or,  respectively,  one-sixteenth  and  one-eighth  of  a  pound  per 
day.  These  figures,  are  of  course,  the  theoretical  figures  for  an 
ampere  efficiency  of  100  per  cent. 

I.     1. — TREATMENT   OF   ORES   BY   ELECTROLYSIS. 

There  are  no  native  ores  of  copper  susceptible  of  being  melted 
and  electrolyzed — in  the  manner  for  instance,  that  sodium  nitrate 
(Chili  saltpeter)  can  be  melted  and  electrolyzed  for  the  pro- 
duction of  sodium.  The  most  abundant  ore  of  copper,  its 


THE  METALLURGY  OF  COPPER.  537 

sulphide,  occurs  mostly  mixed  with  several  times  its  weight 
of  foreign  material,  and  even  if  picked  out  pure  and  melted  it 
redissolves  copper  at  the  cathode  so  actively  that  no  electrolysis 
is  possible.  Faraday  showed  that  melted  cupric  oxide  is  de- 
composed by  the  current ;  and  if  the  oxides  of  copper  were  found 
anywhere  in  sufficient  purity  and  quantity  they  could  be 
melted  and  decomposed  electrolytically,  but  hardly  in  any 
case  as  cheaply  as  they  can  be  reduced  by  carbonaceous 
material. 

Cupric  chloride  is  found  in  nature  as  atacamite,  with  the 
formula  CuCP.  3Cu2O.  2H2O,  in  a  comparatively  pure  condi- 
tion, but  in  small  quantity,  in  Chili.  Such  material  will  dis- 
solve in  considerably  quantity  in  melted  salt  (NaCl),  and  can 
then  be  electrolyzed.  The  dissolved  copper  salt  is  first  re- 
duced by  the  current,  with  evolution  of  chlorine,  to  CuCl,  and 
then  this  decomposed.  The  quantity  of  this  material  is  at 
present  insignificant,  but  there  is  a  possibility  of  an  electrolytic 
process  along  this  line. 

Cuprous  chloride  does  not  occur  in  nature,  but  can  be  readily 
made  from  other  copper-bearing  material.  Chlorine  gas, 
for  instance,  converts  the  common  copper  sulphide,  Cu2S, 
into  CuCl  (Ashcroft's  process).  When  melted,  cuprous  chloride 
conducts  the  current  very  well,  copper  separating  out  as  fine 
leaves.  The  melt  cannot  be  heated  to  the  melting  point  of 
copper  and  the  copper  obtained  liquid,  because  it  vaporizes 
too  easily.  It  is  a  good  conductor,  its  resistivity  at  50°  above 
its  melting  point  being  only  6  ohms  (per  centimeter  cube). 

Problem  111. 

In  electrolyzing  a  bath  of  melted  cuprous  chloride,  using  an 
unattackable  anode,  the  electrodes  are  4  centimeters  apart,  and 
a  current  density  of  0.5  amperes  per  square  centimeter  is  used. 

Required : 

(1)  The  voltage  required  for  running  the  bath. 

(2)  The  proportion  of  the  energy  of  the  current  converted 
into  heat. 

(3)  The  volume  of  chlorine,  at  0°,  liberated  per  minute  by 
a  current  of  1,000  amperes. 

(4)  The  output  of  copper  in  kilograms  per  kilowatt-hour  of 
electric  energy  employed. 


538  METALLURGICAL  CALCULATIONS. 

Solution : 

(1)  The   voltage   required   includes  that  necessary  to   over- 
come the  ohmic  resistance  of  the  electrolyte  plus  that  absorbed 
in  chemical  decomposition.     With  a  resistivity  of  6  ohms,  cur- 
rent density  0.5  amperes  per  square  centimeter,  and  distance  be- 
tween electrodes  4  cm.,  the  first  item  is 

V<  =  6X0.5X4=  12 

The  second  item  comes  from  the  heat  of  formation,  by  divid- 
ing that  heat  expressed  per  chemical  equivalent  by  23,040. 
Since  (Cu,  Cl)  =  35,400  Calories,  we  have 

Vd  =  35,400^23,040  =  1.5 
The  total  voltage  required  would  be 

Vc  +Vd  =  13.5  volts.  (1) 

This  is  independent  of  drop  of  voltage  at  contacts  of  electrodes 
with  the  conductors.  .That  may  amount  to  0.5  or  1.0  volt, 
unless  the  contacts  are  very  closely  looked  after. 

(2)  The  current  drops  as  sensible  heat: 

12  4- 13.5  =  0.89  =  89  per  cent  of  its  energy.  (2) 

(3)  Assuming  100  per  cent  efficiency  of  liberation  of  chlorine, 
we  have: 

Hydrogen  gas  liberated  by  1  ampere,  in  1  sec.=  0.00001036  gr. 
Chlorine  gas  liberated  by  1  ampere,  in  1  sec. 

0.00001036X35.5     (Chem.     eq.     wt.)  =    0.0003677     " 
Chlorine  gas  per  1000  amperes  in  1  minute  =  22.06  " 

Volume  of  chlorine,  at  0°  C. 

22.06 -K0.09X  35.5)  =  6.9  litres.  (3) 

(4)  One  kilowatt,  at  13.5  volts,  gives  1000 -v- 13.5  =  74  am- 
peres.   This  current,  in  one  hour,  will  furnish,  at  100  per  cent 
efficiency 

2.372X74  =  175  grams  of  copper.  (4) 

Lower  current  densities  would  absorb  less  voltage  and  give 
a  greater  power-factor  output. 


THE  METALLURGY  OF  COPPER.  539 

I.     2 — ELECTROLYTIC  TREATMENT  OF  MATTE. 

Matte  is  a  mixture  of  Cu2S  with  FeS,  and  frequently  im- 
pure with  Pb,  Zn,  Ag,  Au,  As,  Sb,  Ba,  etc.  It  melts  sharply 
at  about  1,000°  C.,  to  a  thin  liquid;  it  sets  quickly  to  a  hard, 
stony  mass.  When  melted  it  can  dissolve  copper  rapidly,  so 
that  electrolysis  results  in  the  matte  being  re-formed  as  quickly 
as  it  tends  to  be  decomposed,  and  it  apparently  conducts  the 
current  without  decomposition.  If  it  could  be  dissolved  in 
certain  other  fused  sulphides  in  small  amount,  such  as  in  fused 
sodium  sulphide,  it  could  conceivably  be  electrolyzed  continu- 
ously, but  no  process  has  as  yet  been  developed  along  these  lines. 
If  it  could  be  dissolved  in  aqueous  solutions  of  alkaline  or 
other  sulphides,  it  might  be  electrolyzed  in  such  solvents, 
but  no  successful  solvent  of  this  nature  has  been  found.  It  is 
not  impossible  that  some  aqueous  solutions  may  be  found  to 
answer  this  purpose. 

Solid  matte  is  conducting,  and  may  be  used  as  an  anode  or 
a  cathode.  When  so  used  it  is  acted  upon  by  the  electric  cur- 
rent, reducingly  as  cathode  and  perducingly  as  anode;  that  is, 
used  as  cathode  it  tends  to  be  reduced  in  situ  to  metallic  cop- 
per and  iron,  if  there  is  any  base  present  capable  of  uniting 
with  and  carrying  away  the  sulphur;  used  as  anode  its  copper 
tends  to  pass  into  combination  with  the  acid  radicle  of  the 
electrolyte,  leaving  the  sulphur  behind. 

USE  OF  MATTE  AS  ANODE. 

This  has  appeared  to  many  persons  a  hopeful  application  of 
electrometallurgy  to  copper.  The  matte  is  cast  around  some 
strips  of  copper,  or  copper  netting  is  better,  these  giving  the  piece 
strength,  preventing  its  disintegrating  too  quickly,  and  serving 
as  conductors.  Used  in  acidulated  copper  sulphate  solution, 
the  tendency  is  to  dissolve  out  copper  as  copper  sulphate  and 
leave  the  sulphur  as  a  residue.  The  latter  is  a  non-conductor, 
forming  an  insulating  layer,  which  increases  greatly  the  voltage 
required  to  run  the  bath.  Attempts  to  scrape  off  this  layer  are 
fruitless,  because  of  the  irregular,  cavernous  corrosion  of  the 
anodes.  Marchesi,  an  Italian,  installed  a  plant  to  work  this 
process  near  Genoa  in  1882;  a  plant  was  also  erected  in  Stol- 
berg,  Germany,  Both  were  subsequently  abandoned  as  im- 
rjracticable. 


540  METALLURGICAL  CALCULATIONS. 

The  theoretical  reaction  is  for  purest  matte 
Cu2S  =  2Cu  +  S 

so  that  for  four  chemical  equivalents  of  copper  deposited  one 
molecule  of  Cu2S  is  broken  up.  The  voltage  corresponding  to 
the  chemical  work  done  is,  therefore, 


The  voltage  required  by  the  cell  is  this  plus  that  absorbed  in 
overcoming  the  ohmic  resistances  of  the  circuit. 

If  the  matte  were  only  FeS,  the  reaction  would  consist  in 
the  solution  of  iron,  sliming  of  sulphur,  and  deposition  of  an 
equivalent  amount  of  copper: 

FeS  +  CuSO4aq.  =  Cu  +  FeSO4aq.  +  S 
and  the  energies  involved  are 

(Fe,  S)  =    24,000  Calories  absorbed 

(Cu,  S,  O4,  aq.)  =  197,500      " 

(Fe,  S,  O4,  aq.)  =  234,900      "       evolved 


Sum  =     13,400 

and  the  voltage  contributed  to  the  circuit  is 
13,400  ^  2 


23,040 


=  0.29  volt. 


If  the  matte  is,  as  it  usually  is,  part  Cu2S  and  part  FeS,  then 
when  it  is  uniformly  corroded  there  occurs  a  combination  of 
the  above  two  reactions.  If  the  matte  corresponded,  for  in- 
stance, to  the  formula  Cu2S.FeS,  and  these  were  simultaneously 
acted  upon,  the  current  would  be  divided  in  the  proportions 
required  by  the  preceding  reactions;  that  is,  twice  as  much  to 
handle  the  Cu2S  as  for  the  FeS.  The  voltages  concerned  will 
enter  into  the  calculation  in  the  proportions  2  to  1,  and  the 
calculated  voltage  of  decomposition  will  be 

0.22X|  =       0.147  volt  absorbed. 
0.29  X     =      0.097     "     contributed. 


Sum  =       0.05     "      absorbed. 


THE  METALLURGY  OF  COPPER.  541 

The  calculation  can  be  made  for  any  proportions  of  Cu2S 
and  FeS,  remembering  that  every  159.2  parts  of  Cu2S  require 
twice  as  much  current  as  88  parts  of  FeS. 

Problem  112. 

Marchesi  erected  and  operated  for  some  time  an  electrolytic 
plant  using  copper  matte  as  anodes.  The  matte  used  contained 
30  per  cent  copper,  30  sulphur  and  40  iron,  and  the  anodes 
measured  800  x  800  x  30  mm.,  and  weighed  125  kilograms. 
The  cathodes  were  700  x  700  x  0.3  mm.  The  tanks  were 
lead  lined,  with  interior  dimensions  2,000  x  900  mm.  x  1,000 
mm.  deep.  Electrolyte  an  acid  solution  of  copper  and  iron 
sulphates.  Resistivity  assumed  at  6  ohms.  The  plant  con- 
tained 120  tanks,  arranged  in  ten  groups  of  twelve  each,  each 
group  being  run  by  a  separate  dynamo.  Current  density,  30 
amperes  per  sq.  meter  of  cathode  surface;  each  tank  contained 
twenty  anodes  and  twenty-one  cathodes.  Conductors,  30  mm. 
diameter;  total  length  to  one  group  10  meters. 

Required : 

(1)  The  electrical  current  required  by  each  group  of  twelve 
tanks  and  the  motive  power  to  run  its  dynamo. 

(2)  The  rate  at  which  the  anodes  lost  weight  per  day. 

(3)  The  rate  at  which  FeSO4  accumulates  in  the  bath,  in 
per  cent  of  the  weight  of  the  bath. 

(4)  The  output  of  copper  per  day. 

(5)  The  length  of  time  necessary  to  plate    1  cm.  thickness 
of  copper  on  one  side  of  each  cathode  plate. 

Solution : 

(1)  The  first  point  to  be  solved  is  the  voltage  required  per 
tank,  and  that  consists  of  voltage  absorbed  by  ohmic  resistance 
and  that  of  chemical  decomposition.  The  first  must  be  found 
from  the  ohmic  resistance  of  the  baths,  the  second  from  the 
chemical  reactions  involved. 

With  twenty  anodes,  each  30  mm.  thick,  and  twenty-one 
cathodes  (the  end  ones  against  the  ends  of  the  tank)  0.3  mm. 
thick,  the  thickness  of  electrodes  in  a  tank  is 

(20X30) -f- (21X0.3)  =  606.3  mm. 
=    60.6  cm. 


542  METALLURGICAL  CALCULATIONS. 

and  the  free  space  between,  in  the  length  of  the  tank,  is 

200.0  -  60.6  =  139.4  cm. 
Since  this  is  divided  into  40  spaces,  the  free  space  is 

139.4 -f- 40  =  3.5  cm. 
The  total  anode  surface  is,  assuming  them  entirely  immersed, 

80.0X80.0X2X20  =  256,000  sq.  cm. 
and  of  the  cathodes, 

70.0X70.0X2X20  =  196,000  sq.  cm. 
giving  current  passing  through  a  tank 

19.6X30  =  588  amperes. 

Since  the  tank  is  wider  and  deeper  than  the  plates,  the  ef- 
fective cross-sectional  area  of  electrolyte  is  25.6  sq.  m.  at  the 
anode  surface,  19.6  sq.  m.  at  the  cathode  surface,  and  greater 
than  either  (by  spreading  of  current  lines)  in  between.  It 
will  not  be  far  wrong,  under  these  conditions,  to  take  the  ef- 
fective area  of  the  electrolyte  as  about  that  of  the  larger  elec- 
trode, viz.,  at  25.6  sq.  m.  The  resistance  of  the  tank  thus 
becomes 

R  -  SSo  -  °-000082  °hm- 

and  the  voltage  absorbed  in  overcoming  ohmic  resistance 
0.000082X588  =  0.048  volt. 

During  active  corrosion,  if  copper  and  iron  are  dissolved  in 
the  proportions  30  to  40,  this  would  be,  in  chemical  equivalent 

proportions  as  30  ^  3 1 . 8  to  40  -r-  28 

or  as  0.943  to  1.429 

since  the  current  divides,  in  doing  mixed  electrolysis,  in  pro- 
portion to  the  number  of  chemical  equivalents  dissolved  or 
deposited.  It  thus  results  that  0.943-^(0.943  +  1.429)  =0.4  = 
40  per  cent  of  the  current  is  dissolving  copper  and  60  per  cent 
dissolving  iron.  The  corresponding  voltage  of  the  chemical 
work  is.  therefore* 


THE  METALLURGY  OF  COPPER.  543 

0.22X0.40  =       0.088 
0.29X0.60  =  -0.174 


Sum  =  -  0.086  volt. 
The  sum  of  these  two  voltages  is 

Vc  =       0.048  volt 
V*  =  -  0.086     " 


V  =  -  0.038     " 

The  conclusion  is,  that  as  long  as  the  surface  of  the  matte 
is  clean,  and  no  resistance  offered  by  the  non-conducting  sul- 
phur slime,  the  bath  will  really  require  no  outside  current 
to  run  it,  but  will  practically  run  itself.  The  resistance  of  con- 
nections would  absorb  the  small  excess  of  voltage  produced. 
This  is  as  far  as  calculation  can  go.  Practical  experience  with 
the  bath  records  that  the  voltage  required  quickly  rose  to  1 
volt,  and  after  a  few  days  running  reached  5  volts.  It  is  prac- 
tically certain  that  this  was  caused  entirely  by  the  non-conducting 
film  formed  on  the  anodes. 

(2)  With  588  amps,  passing,  and  40  per  cent  dissolving  cop- 
per, or  235  amps.,  the  copper  dissolved  per  day  in  one  tank  is 

28.46X235  =  6688  grams 

=  6.688  kilograms 

and  the  weight  of  matte  dissolved  per  day 

6.688-^0.30  =  22.3  kilograms. 

Since  the  20  anodes  in  a  tank  weigh  20X125  =  2500  kilograms, 
their  loss  of  weight  per  day  is 

22.3-2500  =  0.009  =  0.9  per  cent  (2) 

and  to  lose  their  whole  weight  would  theoretically  require 

* 

100-;- 0.9  =  111  days. 

Practically,  the  plates  went  to  pieces  in  about  half  that  time, 
when  about  one-half  of  their  weight  had  been  dissolved. 

(3)  The  iron  dissolved  is  40  per  cent  of  the  matte  decom- 
posed, or  8.9  kilograms  per  day.     This  forms 


544  METALLURGICAL  CALCULATIONS. 

1  ^9 

8..9X-T5T  =  24.2kg.  FeSO4. 
oo 

The  tank  is  not  quite  full  of  solution,  but  say  to  within  3  cm.  of 
the  top.     This  gives  a  space  of 

200X90X97  =  1, 746,000  cc. 

1.746  cu.  m. 

From  this  would  be  deducted  the  volume  of  the  electrodes: 

80  X  80  X  3  X  20       =  384,000  cc. 
70X70X0.3X21  =     34,300    " 


Sum  =  418,300    ' 

=       0.418  cu.  m. 
Volume  of  solution  =       1.328    " 

Assuming  its  specific  gravity  as  1.2,  the  weight  of  solution  in  a 
tank  is 

1,328X1.2  =  1594  kilograms 

and  its  content  of  FeSO4  is  increased  per  day 

24.2 -- 1594  =  0.015  =  1.5  per  cent.  (3) 

The  total  salts  present  in  the  electrolyte,  however,  do  not 
increase  quite  that  fast,  for  since  16.73  kg.  of  copper  are  de- 
posited per  day  and  only  6.67  kg.  are  dissolved,  the  electrolyte 
loses  copper  at  the  rate  of  10.06  kg.  per  day,  equal  to  25.24  kg. 
of  CuSO4  per  day,  which  is  1.6  per  cent  of  the  weight  of  the 
electrolyte.  We  may,  therefore,  say  that  the  electrolyte  would 
lose  1.6  per  cent  of  its  weight  of  CuSO4  per  day  and  gain  1.5 
per  cent  of  FeSO4 — amounting  to  a  virtual  displacement  of 
CuSO4  by  FeSO4  until  the  copper  was  all  removed.  This  would 
result  in  not  many  days  running  before  the  electrolyte  would 
have  to  be  replaced. 

(4)  We  have  previously  calculated  for  one  tank,  that  16.73 
kg.  of  copper  is  deposited  per  day.  This  amounts  for  the  whole 
plant  to 

16.73X120  =  2008  kilograms.  (4) 

Of    this    amount,    however,    only    6.67X120  =  800    kg.    came 
from  the  matte,  and  the  rest  from  the  solution.     Ample  means 


THE  METALLURGY  OF  COPPER.  545 

must  therefore  be  provided  to  supply  fresh  CuSO4  solution, 
which  was  obtained  by  Marchesi  from  the  roasting  and  leaching 
of  ore. 

(5)  Precipitated  copper  has  a  density  of  8.9.  One  amp. 
precipitates  per  day  28.46  grams.  One  amp.  per  square  centi- 
meter would  therefore  deposit  in  a  day 

'       =  3.2  cubic  cm. 
o.y 

=  3.2  cm.  thickness. 

But  the  current  density  used  for  depositing  by  Marchesi  was 
only  30  amps,  per  square  meter,  =  0.003  amps.,  per  square 
centimeter,  which  would  deposit  a  layer  per  day  of 

3.2X0.003  =  0.0096  cm. 
=  0.096  mm. 

To  deposit  a  layer  1  cm.  thick  would  therefore  require 

1.0 -i- 0.0096  =  104  + days.  (5) 

This  is  a  much  slower  rate  of  deposition  than  is  used  at  pres- 
ent in  copper  refining,  where  100  to  500  amps,  per  square  meter 
(9  to  45  amps,  per  square  foot)  are  used.  Low-current  density 
deposits  purer  copper  and  absorbs  less  power,  but  gives  a 
smaller  output  for  a  given  installation,  with  consequent  higher 
interest  charges,  labor  costs,  amortisation  and  general  ex- 
penses. 

USE  OF  MATTE  AS  CATHODE. 

Pedro  G.  Salom  describes  the  use  of  this  principle,  called  by 
him  cathodic  reduction,  and  has  applied  it  on  a  large  scale  to 
the  reduction  of  PbS  concentrates,  used  as  cathode  in  dilute 
sulphuric  acid,  to  spongy  lead.  If  the  process  is  applied  to 
granulated  copper  matte,  assumed  as  nearly  pure  Cu2S  to 
simplify  this  discussion,  the  anode  product  is  O2  gas,  and  the 
cathode  product  H2  gas  mixed  with  varying  quantities  of  H2S. 
A.  T.  Weigh tman  (Transactions  American  Electrochemical  So- 
ciety, II  (1902),  p.  76)  gives  details  of  such  an  experiment. 


646  METALLURGICAL  CALCULATIONS. 

Problem  113. 

Fifteen  grams  of  Cu2S  were  used  as  cathode  in  a  5  per  cent 
solution  of  H2SO4,  using  an  unattackable  antimonial-lead 
anode.  A  current  of  3.6  amps,  was  sent  through  for  3  hours 
(voltage  not  given).  Area  of  electrodes  50  sq.  cm.  each,  dis- 
tance apart  4  cm.  Gases  contained  H2  and  H2S  in  the  follow- 
ing proportions: 

H2  H2S 

At      5'  42.4         57.6 

At  180'  90.8  9.2 


Average  0-180'  79.5         20.5 

Required: 

(1)  The  voltage  required  to  run  the  cell,  if  H2S  alone  were 
liberated,  100  per  cent  pure. 

(2)  The  voltage  required  to  run  the  cell  at  the  beginning,  at 
the  end,  and  the  average  throughout  the  run. 

(3)  The  proportion  of  the,Cu2S  reduced  to  Cu  during  the  run. 

Solution : 

(1)  The  voltage  to  overcome  ohmic  resistance  requires  first 
the  resistivity  of  the  electrolyte,  which  for  5  per  cent  solution 
is  4.8  ohms.  The  ohmic  resistance  of  the  cell  is,  therefore, 

4.8X4^-50  =  0.38  ohm 

and  the  voltage  drop  to  send  3.6  amperes  through  this 
0.38X3.6=1.37  volts. 

If  H2S  were  liberated  pure,  in  which  case  the  solution  would 
be  saturated  with  EPS,  the  chemical  reactions  are 

Cu2S  +  H2O  =  2Cu  +  H2S  (gas)+O 
and  the  energy  involved 

(Cu2,  S)  =  20,300  absorbed 

(H2,  O)  =  69,000 

(H2,  S)  =    4,800  evolved 


Sum  =  84,500  absorbed. 

The  voltage  absorbed  in  producing  this  chemical  reaction  is: 
84,500 -h  2 


23.040 


=  1.83  volts 


THE  METALLURGY  OF  COPPER.  547 

The  total  voltage  necessary  to  run  the  cell  is 

1.37+1.83  =  3.20  volts.  (1) 

(2)  If  no  EPS  were  formed,  and  the  current  liberated  only 
H2  gas,  there  would  be  absorbed  in  decomposition  1.50  volts, 
and  by  the  cell  as  a  whole  2.87  volts.  Since  a  molecule  of  H2S 
contains  the  same  amount  of  hydrogen  as  a  molecule  of  H2, 
equal  volumes  of  H2S  or  H2  would  be  liberated  by  an  equal 
electric  current.  It  follows,  therefore,  that  at  5  minutes  57.6 
per  cent  of  the  current  was  performing  reduction,  liberating 
H2S,  and  42.4  per  cent  liberating  hydrogen. 

The  principle  here  involved  in  calculating  the  voltage  dropped 
corresponding  to  the  chemical  work  done  is  that  of  the  com- 
position and  resolution  of  voltages,  explained  by  the  writer 
in  Transactions  American  Electrochemical  Society  V.  (1904), 
p.  89.  The  numerical  result  desired  may  be  reached  in  several 
ways,  for  instance, 

1.83X0.576  =  1.054 
1.50X0.424  =  0.636 


V<*  =  1.69    volts, 
but,  for  conduction,  Vc  =  1.37        " 

therefore,  total  voltage          V  =  3.06        " 

(2) 
At  the  end  of  the  run  we  have  similarly 

1.83X0.092  =  0.168  volts 
1.50X0.908  =  1.362      " 


Vd  =  1.53 
V<  =  1.37 


V  =  2.90 

(2) 

For  the  average  running 

1.83X20.5  =  0.375  volts 
1.50X79.5  =  1.193      " 


Vd  =  1.57 
Vc  =  1.37 

V  =  2.94       "  (2) 


548  METALLURGICAL  CALCULATIONS. 

(3)   Since  20.5  per  cent  of  the  current  reduced  copper  during 
the  run,  the  weight  of  copper  reduced  must  have  been 

1.186X3.6X0.205  =  0.88  grams. 
But,  the  15  grams  Cu2S  contained,  if  pure, 
127  2 


Proportion  reduced  in  the  three  hours 
0.88 


12 


0.073  =  7.3  per  cent.  (3) 


This  method  of  reduction  of  matte  would  need  radical  im- 
provement before  there  could  be  any  possibility  of  its  com- 
mercial application. 

I.     3 — ELECTROLYTIC  EXTRACTION  FROM  SOLUTIONS. 

This  branch  of  the  subject  covers  some  promising  processes 
which  have  not,  however,  been  as  yet  practically  successful. 
Their  consideration,  however,  from  a  quantitative  point  of 
view,  is  not  devoid  of  interest  or  lacking  in  instruction. 

The  waters  of  many  copper  mines  carry  copper  sulphate, 
produced  from  the  weathering  and  leaching  of  copper  sulphide 
ores.  The  solutions  are  generally  too  dilute  to  allow  of 
concentration  and  crystallizing  out  to  blue  vitriol.  The  stand- 
ard method  of  treatment  is  to  pass  the  solutions  over  pig  iron 
or  scrap  iron,  thus  precipitating  the  copper  as  a  sort  of  metallic 
mud  called  "  cement  copper,"  which  contains  much  iron,  be- 
sides the  impurities  of  the  iron  used,  so  that  it  is  sometimes 
only  90  per  cent  down  to  60  per  cent  copper.  This  precipitate 
needs  a  strong  refining  to  bring  it  up  to  merchant  copper,  with 
considerable  loss  of  copper  in  the  operation. 

Two  electrolytic  methods  are  applicable  to  the  treatment  of 
such  solutions:  1.  The  use  of  soluble  iron  anodes.  2.  The  use 
of  insoluble  anodes. 

(1).      Use  of  Soluble  Iron  Anodes. 

If  iron  in  plates  or  sheets,  or  bundles  of  scrap  iron  in  a  crate 
or  holder  are  immersed  in  copper  sulphate  solution  and  simul- 
taneously connected  electrically  with  a  copper  plate  to  serve 


THE  METALLURGY  OF  COPPER.  549 

as  cathode,  no  copper  precipitates  on  the  iron,  but  all  is  pre- 
cipated  on  the  copper.  The  iron  acts  as  a  soluble  anode,  going 
into  solution  as  ferrous  sulphate,  while  the  copper  is  deposited 
out  passive,  dense,  and  practically  chemically  pure,  on  the 
cathode  sheet.  Since  more  energy  is  developed  by  the  solution 
of  the  iron  than  is  absorbed  in  the  deposition  of  the  copper, 
there  is  electromotive  force  generated  by  the  chemical  action, 
which  will  run  the  tank  like  a  short-circuited  battery  cell,  if 
the  iron  and  copper  are  simply  connected  by  a  low  resistance 
wire.  This  electromotive  force  will  send  a  current  of  a  certain 
amount  through  the  cell,  depending  on  its  internal  ohmic  resis- 
tance plus  the  resistance  of  the  external  conductor.  If  it  ig 
desired  to  force  matters,  and  to  precipitate  the  copper  faster 
than  this  auto-precipitation,  an  impressed  electromotive  force/ 
from  a  dynamo  may  be  put  on  and  the  cell  made  to  work  faster. 

Problem  114. 

A  copper  sulphate  solution  whose  resistivity  is  50  ohms  is 
run  for  precipitation  through  tanks,  each  of  which  contains 
fifteen  anodes  of  cast  iron  40  x  80  cm.  in  size,  and  sixteen 
sheets  of  copper  of  similar  size,  the  distance  between  aver- 
aging 5  cm.  The  anodes  and  cathodes  are  short-circuited  by 
resting  on  a  triangular  copper  distributing  bar  of  negligible 
resistance.  Assume  resistance  of  contacts  to  be  such  that  0.1 
volt  will  be  lost  at  them. 

Required : 

(1)  The  electromotive  force  generated  by  the  chemical  action. 

(2)  The  total  current  operative  in  each  tank,  and  the  current 
density. 

(3)  The  weight  of  copper  deposited  in  each  tank  per  day. 

(4)  The  weight  of  iron  dissolved  in  each  tank  per  day. 
Solution : 

(1)  From  the  thermochemical  tables  (Metallurgical  Cal- 
culations, Part  I.  p.  24)  we  have: 

(Fe,  S,  O4,  aq.)  =  234,900  Calories 

(Cu,  S,  O4  aq.)  =  197,500 

Excess  of  anode  energy  =    37,400 

Since  this  is  generated  for  each  atom  of  copper  deposited 
and  of  iron  dissolved,  the  excess  energy  per  chemical  equiva- 


550  METALLURGICAL  CALCULATIONS. 

lent  concerned  is  37,400-2  =  18,700  Calories,   and  the  total 
electromotive  force  developed  is 

18,700-23,040  =  0.81  volt.  (1) 

(2)  The  loss  of  voltage  at  the  contacts  being  0.1  volt,  there 
is  0.71  volt  operative  to  overcome  the  ohmic  resistance  of  the 
solution.     From  the  data  given,  the  fifteen  anodes  operating 
on  both  sides,  sandwiched  between  sixteen  cathodes,  give 

15X2X40X80  =  96,000  sq.  cm. 

active  electrode  surface,   and  taking  this  as  the  cross-section 
of  the  electrolyte  between,  we  have  its  resistance  as 

50 -=-96,000X5  =  0.0026  ohm, 
the  total  current  passing  in  the  tank 

0.71-0.0026  =  273  amperes,  (2) 

and  the  current  density 

273  —  9.6  =  28.4  amperes  per  square  meter  (2) 

=    2.6  amperes  per  square  foot. 

(3)  Copper  deposited  in  the  tank  per  day: 

28.46X273  =   7770  grams 

=     7.77  kilograms 

=  17.13  pounds.  (3) 

(4)  The  iron  going  into  solution  will  be  to  the  copper  dis- 
solved as  56  to  63.6.     If  the  iron  is  cast  iron,  this  may  be  only 
90  to  93  per  cent  of  the  loss  of  weight  of  the  anodes,  because 
they  contain  only  that  percentage  of  iron.     If  wrought  iron  or 
steel   sheets   are   used,  the   iron  dissolved   would  represent   99 
to  99.8  per  cent  of  the  loss  of  weight  of  the  anodes.     The  iron 
dissolved  per  day  is 

7 . 77  X  56  -  63 . 6  =    6.84  kilograms  (4) 

=  15.08  pounds. 

Problem  115. 

One  hundred  tanks  of  the  kind  described  in  Problem  114  are 
arranged  in  a  series,  the  anodes  in  each  tank,  however,  con- 
nected as  one  large  anode,  and  the  cathodes  in  each  as  one 


THE  METALLURGY  OF  COPPER.  551 

large  cathode.  A  dynamo  is  connected  to  the  series  capable 
of  maintaining  110  volts  across  its  terminals.  The  bus-bars 
provided  are  altogether  280  meters  long,  and  are  1x4  cm.  in 
cross-section,  of  pure  copper.  All  other  details  of  the  tanks 
are  as  before:  resistance  of  contacts  being  0.1  volt -j- 273  amps.  = 
0.0004  ohm. 
Required : 

(1)  The  ohmic  resistance  of  the  bus-bars. 

(2)  The   current   operative   in   the   circuit,   and   the   current 
density. 

(3)  The  weight  of  copper  deposited  in  each  tank  per  day. 
Solution : 

(1)  The  conductivity  of  copper,  in  reciprocal  ohms,  is  600,000; 
its    resistivity     1-i- 600,000  =  0.00000167    ohms.     This    is    its 
resistance  per  centimeter  cube.     The  resistance  of  the  bus-bars 
is  therefore 

0 . 00000167  X  28000  +  4  =  0.0117  ohm.  (1) 

(2)  The  total  voltage  operative  in  the  circuit  is  the  self- gen- 
erated voltage  of  100  tanks,  plus  the  110  volts  from  the  dynamo, 
or 

(0.81X100) +  110  =  191  volts. 

The  total  resistance  of  the  circuit  is  that  of  the  100  tanks, 
plus  100  sets  of  connections,  plus  that  of  the  bus-bars 

(0.0026X100) +  (0.0004X100) +0.01 17  =  0.3117  ohm.    • 
The  total  current  flowing  will  therefore  be 

191 -T- 0.31 17  =  618  amperes  (2) 

and  the  current  density 

618-7-9.6   =  64.4  amperes  per  square  meter     (2) 
=     5.9  amperes  per  square  foot. 

(3)  28.46X618  -*- 1000  =  17.59  kilograms  per  day  (3) 

=  38.79  pounds  per  day. 

(2).   Use  of  Insoluble  Anodes. 

If  insoluble  anodes  are  used,  the  copper  may  be  extracted 
en  masse  from  such  solutions,  and  its  acid  left  behind.  In 
such  operations,  the  choice  of  an  anode  is  not  easy.  Graph- 


552  METALLURGICAL  CALCULATIONS. 

itized  carbon  plates  are  unattacked,  and  have  practically  no 
back  electromotive  force;  high-silicon  iron  plates  are  also  said 
to  be  very  resistant,  and  pure  silicon  itself  resists  still  better, 
but  introduces  considerable  ohmic  resistance.  Sheet  lead  an- 
odes become  coated  with  a  brown  coating  of  PbO2,  which  is 
permanent  and  evolves  oxygen  freely.  Many  other  substances 
may  possibly  be  used  as  anodes;  not  many  practical  results  of 
such  tests  have  been  published.  When  operating  thus,  the 
electrolyte  offers  practically  the  same  ohmic  resistance  as  be- 
fore, averaging  probably  a  little  less,  and  the  chemical  reac- 
tion absorbs  energy.  The  resulting  acidified  solution  may  be 
utilized  for  dissolving  easily-attacked  copper  compounds  from 
fresh  quantities  of  raw  or  roasted  ore. 

Problem  116. 

An  electrolytic  depositing  plant  treats  dilute  copper  sulphate 
solution  containing  318  grams  of  copper  per  cubic  meter  as 
CuSO4,  and  its  copper  is  deposited  by  passing  through  tanks 
having  insoluble  anodes,  electrodes  3  centimeters  apart,  and 
current  density  of  20  amps,  per  square  meter.  With  good  cir- 
culation, copper  is  precipitated  with  no  evolution  of  hydrogen 
until  the  precipitation  is  practically  complete.  By  having  many 
tanks  in  series  the  current  passing  is  kept  very  nearly  constant 
at  300  amps.  Each  tank  contains  1.5  cu.  m.  of  electrolyte. 

Required : 

(1)  The   voltage    absorbed   in   the   chemical   reaction   when 
precipitating  copper,  and  when  the  copper  is  all  precipitated. 

(2)  The  voltage  absorbed  in  overcoming  the  ohmic  resistance 
of  the  electrolyte  at  starting  and  at  the  end  of  the  precipitation. 

(3)  The   total  voltage   across  the   electrodes  of  a  tank,   at 
the  beginning  of  the  precipitation,  just  before  the  last  of  the 
copper  is  precipitated  and  after  all  copper  is  precipitated. 

(4)  The  time  required  to  precipitate  the  copper  from  a  batch 
of  solution. 

(5)  The    output    of    copper    per    average    kilowatt-hour    of 
electric  energy  used,  adding  0.2  volts  to  the  potential  across 
the  electrodes  for  loss  in  contacts  and  bus-bars  per  tank. 

Solution : 

(1)  When  precipitating  copper,  we  destroy  CuSO4aq.  and 
form  the  corresponding  H2SO4aq.  Their  heats  of  formation 
being 


THE  METALLURGY  OF  COPPER.  553 

(Cu,  S,  O4,  aq.)    =  197,500  Calories 
(H2,  S,  O4,  aq.)    =  210,200 

and  since  the  reaction  is 

CuSO4aq.  +  H2O  =  H2SO4aq.  +  O  +  Cu 
we  have  to  supply 

(Cu,  S,  O4,  aq.)    =  197,500  Calories 
(H2,  O)  =    69,000 


Total  =  266,500 
and  get  back  (H2,  S,  O4,  aq.)  =  210,200 


therefore  supplying  net          =     56,300 
and  voltage  absorbed  in  decomposition 
56,300-2 


23,040 


=  1.22  volts.  (1) 


When  the  copper  is  all  deposited,  then  only  H2  and  O  appear 
at  the  electrodes,  so  that  the  voltage  absorbed  in  decomposition 
becomes 

69,000-2 


23,040 
(2)  At  starting,  the  solution  contains 

318  X  -——-==  798  grams 

of  CuSO4  per  cubic  meter,  and  since  the  cubic  meter  weighs  at 
15°  practically  the  same  as  water,  i.e.,  999.1  kg.,  our  solution  is 

798-1000     -      nQ  cr.4 

—  -  —  =  0  .  08  per  cent  CuSO4 


and  also  equals  0.798^ —     '- —  =  0.01  normal. 


At  the  end  of  the  electrolysis  the  0.01  normal  CuSO4  solution 
becomes  a  0.01  normal  H2SO4  solution.  The  resistivities  of 
both  these  solutions  can  be  found  at  once  from  electrochemical 


554  METALLURGICAL  CALCULATIONS. 

tables    (e.g.,    Kohlrausch,    in    Landolt-Bornstein-Meyerhoffer's 
Tabellen). 

With  a  current  density  of  20  amps,  per  square  meter  (0.002 
per  square  centimeter)  and  a  working  distance  of  3  cm.,  the 
corresponding  voltages  absorbed  in  overcoming  these  ohmic 
resistances  will  be 

1395X3X0.002  =  8.37  volts 
325X3X0.002  =  1.95      "  (2) 

(3)  The   total   voltage   across   the   electrodes,    assuming   the 
solution  to  contain  no  free  acid  at  starting,  will  be 

at  start,  precipitating  copper:          8.37+1.22  =  9.59  volts 
at  close,  precipitating  last  copper:  1.95+1.22  =  3.17      " 
at  close,  evolving  hydrogen:  1.95+1.50  =  3.45      " 

(4)  Copper  present  in  a  tank:       318x1.5    =  477  grams. 
Time  required,  300  amperes 

477 H- (300X0. 0003295)  =  4826  seconds 

=  1  hr.  20.5  min.    (4) 

(4)  The  average  voltage  consumed  in  overcoming  ohmic 
resistance  will  be  considerably  less  than  the  mean  of  8.37  and 
1.95  =  5.16  volts,  because  when  the  electrolysis  is  only  half 
completed,  and  the  solution  is  0.005  normal  in  each  ingredient, 
its  resistance  will  be  487  ohms,  and  not  (1,395  +  325)^2  = 
860  ohms,  and  the  corresponding  voltage  2.92.  This  is  because 
sulphuric  acid  is  a  so  much  better  conductor  than  CuSO4,  that 
as  soon  as  acid  forms  the  solution  becomes  much  better  con- 
ducting. If  we  calculate  the  resistances  and  corresponding 
voltages  absorbed  in  several  steps,  we  would  get  more  ac- 
curate results.  Without  going  into  the  detailed  calculation  we 
will  take  3.52  as  the  average  voltage,  and  we  then  have  the 
average  voltage  required  per  tank: 

Decomposition    1 . 22  volts 
Resistance  3 . 52      " 

Contacts  0 . 20      " 


Sum  4.94 

Kilowatts  used  per  tank 

4.94X300 


THE  METALLURGY  OF  COPPER.  555 

Output  per  tank  per  hour 

300X1.186  =  355. 8  grams. 
Output  per  kilowatt-hour 

355 . 8  -f- 1 . 482  =  240  grams.  (4) 

This  is,  of  course,  a  small  output  compared  with  that  of  a 
refining  process,  but  it  must  be  remembered  that  this  is  an 
extraction  process,  and  its  cost  is  not  properly  comparable  with 
simple  refining  but  with  that  of  the  processes  which  it  replaces, 
such  as  precipitation  by  iron,  and  with  cheap  power  and  ex- 
pensive iron  there  may  be  localities  where  this  method  would  be 
economical. 

Siemens  and  Halske  invented  a  combined  wet-extraction  and 
electrolytic  process,  consisting  in  leaching  the  roasted  copper 
ore  or  matte  with  a  solution  of  ferric  sulphate  acidulated  with 
sulphuric  acid,  producing  thus  a  solution  of  cupric  and  ferrous 
sulphates,  and  then  electrolyzing  this  solution  in  a  cell  having 
an  insoluble  carbon  anode  and  a  diaphragm  between  the  elec- 
trodes. The  solution  is  run  first  through  the  series  of  cathode 
compartments,  where  its  copper  is  extracted,  and  back  through 
the  anode  compartments,  where  its  ferrous  sulphate  is  per- 
duced  to  ferric  sulphate,  ready  to  be  used  again  in  leaching  ore. 

The  chemical  reactions  in  leaching  ore  are 

Cu2S  +  2Fe2  (SO4)  3aq.  =  2CuSO4aq.  +  4FeSO4aq.  +  S. 
3CuO  +  Fe2(SO4)3aq.  =  3CuSO4aq.  +  Fe2O3. 
CuO  +  H2SO4aq.  =  CuS04aq.  +  H2O. 

In  the  electrical  precipitation  we  have: 

in  the  cathode  compartments:  CuSO4  =  Cu  +  SO4 

in  the  anode  compartments:     2FeSO4  +  SO4        =  Fe2(SO4)3 
altogether:  2FeSO4  +  CuSO4  =  Cu  +  Fe2(SO4)3 

Confining  our  calculations  to  the  electrolysis,  we  have  CuSO4 
destroyed  and  2FeSO4  simultaneously,  perduced  to  Fe2(SO4)8, 
which  is  then  available  for  re-use  in  the  leaching  tanks.  The 
chemical  work  involved  is: 

(Cu,  S,  O4,  aq.)  =  197,500  Calories  absorbed 

2(Fe,  S,  O4,  aq.)  =  469,800 

(Fe2,  S3,  O12,  aq.)         =  650,500        "        evolved 

Sum  =     16,800        "        absorbed 


556  METALLURGICAL  CALCULATIONS. 

Voltage  thus  absorbed  =  16'*°®*2    =  0.36  volt. 


In  addition  to  this,  the  voltage  necessary  to  overcome  the 
ohmic  resistance  of  the  cell  must  be  counted  in. 

Problem  117. 

An  electrolytic  tank  for  working  the  Siemens-  Halske  method 
was  of  wood,  lead  lined,  220  cm.  long,  100  cm.  wide  and  100 
cm.  deep.  Each  cell  contained  fifteen  anodes  and  sixteen 
cathodes,  the  latter  80  x  80  cm.  x  1  mm.  The  anodes  were 
of  carbon  rods  arranged  as  a  grid  45  cm.  x  100  cm.  x  1  cm. 
thick.  A  porous  partition  1  cm.  thick  was  used,  the  equivalent 
effective  free  area  of  which  was  0.05.  Resistivity  of  electrolyte 
m  cathode  compartment  5  ohms,  in  anode  compartment  8 
ohms,  partition  in  middle.  Current  density  100  amperes  per 
square  meter. 

Required  : 

(1)  The   distance   between   the   diaphragm   and  each  anode 
and  cathode  surface,  and  the  resistance  of  the  cell,  taking  into 
consideration  the  contraction  of  current  path  due  to  the  dia- 
phragm. 

(2)  The  voltage  necessary  to  run  the  tank. 

Solution  : 

(1)  15  anodes  XI  =   15.0cm. 

16  cathodes  X  0.1     =      1.6     " 

Sum  =  16.6    " 
Space,  anode  to  cathode 

«  -'.«. 


Space,  anode  or  cathode,  to  diaphragm 

^2^  =  2.9cm.  (1) 

Assume  the  solution  in  the  diaphragm  to  have  mean  re- 
sistivity of  6.5  ohms,  the  equivalent  free  area  being  100  X 
95X0.05  =  475  sq.  cm.  (area  of  diaphragm  a  trifle  less  than 
cross-section  of  tank).  Mean  area  of  electrolyte 


THE  METALLURGY  OF  COPPER.  557 

cathode  compartment  (6400  +  9500)  -j-  2  =  7950  sq.  cm. 

anode  compartment  (4500  +  9500)  H- 2  =  7000    "      " 

Resistance  of 

anode  compartment  8X2.9-7-7000        =  0.0033  ohm. 

cathode  compartment  5X2.9-7-7950        =0.0018      " 

diaphragm  6.5x1-^475          =0.0137      " 


0.0188      u-          (1) 

Voltage  absorbed,  for  64  amperes  (to  0.64  sq.  m.  depositing 
surface) 

0.0188X64  =  1.20  volts 
Vd  =  0.36      " 


1.56      «  (2) 


Carl  Hoepfner  devised  a  process  along  similar  lines  to  the 
above,  but  based  on  the  use  of  the  two  chlorides  of  copper.  A 
solution  of  CuCP  in  brine  is  used  to  act  upon  the  ores  or  roasted 
matte,  dissolving  the  copper  with  the  formation  of  CuCl. 

2CuCP  +  Cu2S  =  4CuCl  + S. 

The  cuprous  chloride  stays  in  solution  in  the  brine,  and  is  elec- 
trolyzed  in  tanks  containing  diaphragms,  half  the  solution 
going  through  the  cathode  compartments,  and  half  through  the 
anode  compartments.  In  the  former,  copper  is  precipitated; 
in  the  latter,  the  equivalent  amount  of  chlorine  converts  the 
CuCl  present  into  CuCP,  which  can  be  used  over,  mixed  with 
the  depleted  cathode  solution.  The  total  reaction  is 

CuCl  +  CuCl  =  Cu  +  CuCP 
and  the  energy  involved 

2(Cu,  Cl)       =  70,800  absorbed 
(Cu,  CP)       =  62,500  evolved 

Sum  =    8,300  absorbed 
=    0.18  volt. 

Coehn  devised  a  cell  for  carrying  on  this  process  without  a 
diaphragm.  The  cathode  is  only  half  as  long  as  the  carbon 
anode,  and  the  CuCP  formed  at  the  latter  is  so  heavy  that  it 
sinks  to  the  bottom  and  is  drawn  off  therefrom,  while  fresh 
CuCl  solution  is  quietly  poured  in  above.  Since  the  cathode 


558  METALLURGICAL  CALCULATIONS. 

does  not  touch  the  cupric  solution,  it  is  not  redissolved  thereby, 
and  the  two  solutions  are  kept  separate. 

A  great  advantage  of  this  process  is  that  Cu  is  monovalent 
in  CuCl,  and  therefore  twice  as  much  copper  is  deposited  as 
from  CuSO4  by  a  given  number  of  amperes. 

Problem  118. 

Coehn  (Jahrbuch  der  Electrochemie,  1895,  p.  155)  describes 
his  improved  partitionless  Hoepfner  apparatus  as  containing  a 
copper  cathode  100  cm.  wi'de  by  50  cm.  deep,  at  a  distance  of 
12  cm.  from  a  carbon  anode;  solution,  10  per  cent  NaCl,  plus 
varying  amounts  of  CuCl  and  CuCP.  Current  density  20  amps. 
per  square  meter. 
Required  : 

(1)  The  voltage  required  to  run  the  cell. 

(2)  The  weight  of  copper  deposited  therein  per  day. 

(3)  The  weight  of  copper  per  kilowatt  hour  electric  power 
used. 

Solution  : 

(1)  The  resistivity  of  10  per  cent  NaCl  solution  is  8.5  ohms. 


Vc  ,for  ohmic  resistance  =  0.0204X10=  0.20  volt 
Vd,  for  chemical  reactions  0.18      " 

V  =  Sum  =  0.38      "  (1) 

(2)  28.46X2X10=    569  grams  per  day        (2) 

(3)  569         1000       =6210  grams 

24       10X0.38   =  6.21  kg.  ^3) 

I.  4.  ELECTROLYTIC  REFINING  OF  IMPURE  COPPER. 
This  subject  is  a  large  one.  It  lends  itself  very  well  to  cal- 
culation. It  is  also  comparatively  simple  in  principle.  Given 
a  nearly  pure  copper  plate  or  slab  as  an  anode,  in  a  solution  of 
a  copper  salt,  and  a  suitable  conducting  cathode,  the  metal  is 
simply  dissolved  and  deposited,  or  is  "  plated  over."  The  pro- 
cess really  consists  in  extracting  copper  from  the  solution  at 
the  cathode,  and  sending  it  and  other  soluble  impurities  into 
the  solution  at  the  anode.  The  mechanical  transportation  of 
the  copper,  in  space,  through  a  distance  equal  to  the  space  be- 
tween anode  and  cathode,  is  sometimes  emphasized  as  the  chief 


THE  METALLURGY  OF  COPPER.  559 

mechanical  work  done.  It  is  a  real  fact  that  the  metal  is  taken 
out  of  the  solution  at  a  different  spot  from  where  it  goes  in, 
but  the  electric  current  does  not  do  that  transportation — dif- 
fusion and  circulation  are  the  agents  which  transport  the  copper 
entering  the  electrolyte  at  the  surface  of  the  anode  plate  to 
the  surface  of  the  cathode  plate,  and  electric  energy  does  nothing 
more  than  create  the  differences  of  concentration,  which  are 
thus  physically  neutralized. 

The  chief  expenditure  of  electrical  energy  in  the  refining 
bath  is  that  converted  into  sensible  heat  in  overcoming  the 
ohmic  resistance  of  the  electrolyte.  The  other  items  of  elec- 
trical work  are  the  overcoming  of  the  ohmic  resistance  of  hangers, 
rods,  bus-bars,  clamps  and  connectors,  and  the  not  unimportant 
resistance  of  contacts;  also  the  difference  of  chemical  work 
done  in  depositing  all  copper  and  dissolving  part  copper  and 
part  other  impurities;  also  the  chemical  work  of  separating 
the  ingredients  of  the  impure  copper,  which  may  be  alloyed 
and  have  some  heat  of  combination  which  must  be  furnished 
by  the  current  in  order  to  separate  them. 

Energy  Absorbed  by  the  Electrolyte. 

The  electrolyte  is  a  conductor  and  absorbs  energy  according 
to  Ohm's  law  whenever  current  passes  through  it.  The  solu- 
tion used  in  copper  refining  is  almost  invariably  copper  sulphate 
acidulated  with  sulphuric  acid.  The  resistivity  of  copper  sul- 
phate, iron  sulphate  and  sulphuric  acid  solutions,  at  20°  C.,  is 
as  follows,  in  ohms  per  centimeter  cube  and  ohms  per  inch 
cube: 

Per  Cent            CuSO4.                    FeSO*.  H2S04. 

in  Q  per  Q  per  Q  per  Q  per       Q  per       Q  per 

Solution.  cm*  inch3.  cm3.  inch3.        cm3.         inch3. 

2.5  92  37             ..              

5.  53  21             ..  ..  4.8           1.9 

7.5  ..  ..  65             26 

10.  31  12             ..  ..  2.5           1.0 

15.  24  10  34  14           1.8          0.7 

17.5  22  9             ..              

20.  .,  ..  ...  ..            1.5           0.6 

25.  ..  .,.              ..  ..  1.4           0.56 

30.  ..  ..  25  10           1.37         0.55 


560  METALLURGICAL  CALCULATIONS. 

The  plain  copper  sulphate  solution  is  thus  seen  to  be  a  rather 
poor  conductor,  iron  sulphate  is  not  so  good;  sulphuric  acid  is 
highly  conducting.  It  is  therefore  evident  that  the  ohmic  re- 
sistance of  the  bath  is,  for  practical  purposes,  dependent  upon 
its  content  in  free  acid;  hence  the  great  economic  importance 
of  keeping  the  bath  well  acidulated.  As  the  bath  grows  foul 
with  iron  sulphate  and  loses  sulphuric  acid  by  formation  of 
insoluble  sulphates,  chemical  solution  of  slimes,  etc.,  the  bath 
becomes  poorer  conducting  and  the  energy  absorbed  in  it  much 
greater.  A  solution  in  good  condition,  with  15  to  20  per  cent 
of  CuSO4  and  5  to  10  per  cent  free  H2SO4,  may  have  a  resist- 
ivity of  only  2  to  5  ohms  at  starting,  which  may  increase  to 
20  or  25  ohms  as  the  solution  becomes  impure  and  the  free  acid 
disappears.  These  facts  are  of  immense  importance  in  the  eco- 
nomics of  refining. 

A  comprehensive,  systematic  study  of  the  resistivities  of 
solutions  of  various  strengths  of  copper  sulphate  with  various 
percentages  of  iron  sulphate  present  and  various  content  of 
sulphuric  acid  is  greatly  needed,  and  would  not  be  a  difficult 
research  to  carry  out. 

Problem  119. 

A  refining  bath  is  run  with  a  current  density  of  250  amperes 
per  square  meter,  with  electrodes  4  cm.  apart,  and  starting  with 
electrolyte  15  per  cent  CuSO4  and  10  per  cent  H2SO4.  It  is 
run  until  the  sulphuric  acid  has  decreased  to  5  per  cent. 

Required : 

(1)  The  probable  voltage  drop  across  the  electrodes  at  start- 
ing. 

(2)  The  same  when  the  second  condition  is  reached. 

(3)  The  rate  at  which  the  solution  would  start  to  rise  in 
temperature  at  the  beginning,  ignoring  the  electrodes. 

(4)  The  rate,  assuming  anodes  5  cm.  thick  and  copper  cath- 
odes 0.5  cm.  thick. 

Solution : 

(1)  Lacking  the  experimentally  determined  datum  of  the 
resistivity  of  the  primary  solution,  we  know  that  it  will  be 
somewhere  about  that  of  the  10  per  cent  sulphuric  acid  solution, 
viz.:  2.5  ohms.  It  should  be  slightly  less,  because  of  the  copper 
salt  present,  and  if  we  assume  the  conductivity  of  the  solutions 


THE  METALLURGY  OF  COPPER.  561 

as  additive,  we  would  have  a  conductivity  of  the  mixed  solu- 
tion, in  reciprocal  ohms  as 

C      -  -L+  *   -  0.40  +  0.04  =  0.44 

_  .  O      ^4 


R5/>  =  =  2.3  ohms. 

Using  this  resistivity,  the  voltage  drop  across  the  electrodes, 
at  starting,  will  be 

Vc  =  2.3X4X0.0250  =  0.23  volts.  (1) 

(0.0250  amperes  goes  across  each  square  centimeter  and  the 
plates  are  4  centimeters  apart.) 

(2)  With  H2SO4  only  5  per  cent,  and  copper  contents,  say, 
20  per  cent,  because  of  solution  of  copper  by  free  acid,  we  would 
have 

~  =  0.21  +  0.05  =  0.26 


Using  this  resistivity,  the  voltage  drop  is 

V*  =  3.6X4X0.0250  =  0.36  volts.  (2) 

In  actual  practice  an  increase  of  0.01  volt  above  this  might  be 
expected,  because  of  the  formation  of  a  film  of  slime  on  the 
surface  of  the  anode. 

(3)  The  solution  at  starting  has  a  specific  gravity  of  1.20 
and  a  specific  heat  of  0.875.  (Combination  of  data  from 
Landolt-Bornstein-Meyerhoffer's  Tabellen.)  The  water  value 
as  heat  absorber  of  a  column  of  liquid  1  cm.2  and  4  cm.  long, 
between  the  electrodes,  is  therefore 

1.20X4X0.875  =  4.2  calories  per  1°  C. 
Heat  equivalent  of  the  electric  energy  expended 

0.23X0.0250X0.2389  =  0.00138  calories  per  1". 


562  METALLURGICAL  CALCULATIONS. 

Rate  of  rise  of  temperature  of  solution 

0 . 00138  -5-  4 . 2  =  0.00033°  per  second 
=  0.0198°  "  minute 
=  1.2°  "  hour  (3) 

=  28.8°  "    day 

(4)  With  half  the  thickness  of  anode  and  cathode  to  be  sup- 
plied with  heat,  the  copper  having  a  specific  gravity  of  8.9  and 
specific  heat  of  0.093,  the  water  value  of  the  copper  concerned 
per  square  centimeter  area  of  electrolyte  section  is 

(-^-  +  -^-)x  8.9X0.093  =  2.276  calories  per  1° 
\   2i          2i   / 

Rate  of  rise  of  temperature  of  solution  plus  electrodes: 

0.00138 ^-(2. 276 +  4. 2)  =     0.00021°  per  second 

=     0.013°        "    minute 
=    0.78°          "    hour 
=  18.7°  "    day  (4) 

Energy  Lost  in  Contacts. 

This  is  an  extremely  variable  and  yet  important  factor  in  the 
economics  of  copper  refining.  No  one  item  in  copper  refining 
will  at  the  present  time  better  repay  close  attention  and  study 
of  methods  of  improvement.  A  proper  contact  should  cause  a 
very  small  voltage  drop  to  begin  with,  and  should  be  capable 
of  being  kept  efficient — that  is  the  chief  requirement.  Con- 
tacts are  often  made  on  the  hit  or  miss  style,  and  arranged  in 
such  position  as  to  receive  spattering  or  drippings  from  the 
electrolyte;  such  may  cause  immense  losses  of  electrical  energy 
as  they  rapidly  pass  from  bad  to  worse. 

B.  Magnus  (Electrochemical  and  Metallurgical  Industry,  De- 
cember, 1903),  and  L.  Addicks  (idem.,  January,  1904)  have 
given  us  valuable  information  on  this  point.  The  weight  of 
the  plates, when  resting  on  the  contacts, improves  the  contact; 
in  such  cases  the  anode  contacts  are  best  when  the  tank  is  first 
set  up  and  the  cathode  contacts  worst.  In  such  cases  a  screw 
clamp,  to  put  several  hundred  pounds  of  mechanically  applied 
pressure  on  the  joint,  should  be  adopted,  so  as  to  make  the 
contacts  "  best  "  all  the  time. 


THE  METALLURGY  OF  COPPER.  563 

Figures  given  by  Magnus  as  a  carefully  determined  average 
of  results  in  a  western  refinery  are: 

Current 4,000  amperes 

Current  density  (per  sq.  ft.). .     11 

Total  voltage  per  tank 0.230    volts 

Energy  used  per  tank 0.920    kw.  Per 


Drop  bus-bar  to  anode  rod.      0.0270  volts 


Cent. 


anode   rod   to   anode  _  Q  QOQQ  _          147 

hanger 0.0060 

"      anode  hook  to  anode.      0.0008 

"      anode  to  cathode. . . .     0.1782       "         =  0.1782  =       77.5 

"      cathode    to    cathode 

rod 0.0045       "     J 

"      cathode    rod  to  bus-  L   =  0.0180  =         7.8 

bar..  0.0135       " 


0.2300          100.0 

It  thus  appears  that  in  this  tank  the  loss  in  contacts  was 
some  20  per  cent  of  the  total  voltage  employed  (assuming  the 
anode  rods  to  be  of  such  design  that  their  resistance  was  neg- 
ligible). Magnus  fitted  up  a  similar  tank  with  mercury  cup 
contacts,  and  found  a  drop  from  bus-bar  to  anode  rod  of  0.0050 
volt  instead  of  0.0270,  and  cathode  rod  to  bus-bar  0.0050  instead 
of  0.0135  volt.  If  it  were  possible  in  the  case  of  above  tank  to 
make  these  improvements  alone,  the  saving  on  voltage  per  tank 
would  be  0.0305  volt,  or  13.3  per  cent  of  the  voltage  used, 
which  means  a  saving  of  13.3  per  cent  of  the  power  required  to 
run  the  tank. 

Problem  120. 

In  an  electrolytic  refinery  system  of  200  tanks  the  resistance 
of  the  contacts  was,  on  an  average,  0.0368  volt  per  tank,  and 
0.2567  volt  across  two  electrodes;  3,800  amperes  was  passed 
through  the  series,  and  the  voltage  at  the  terminals  of  the 
dynamo  was  67  volts.  Power  costs  $157  per  kilowatt-year, 
Efficiency  of  deposition  of  copper  82  per  cent. 
Required : 

(1)  The   cost   of  power  per  ton    (2,000   pounds)    of  copper 
refined. 

(2)  The  cost  per  ton  of  copper  produced,  of  the  power  lost 


564  METALLURGICAL  CALCULATIONS. 

by  (a) ,  resistance  of  the  conductor  bars  (b) ,  by  the  contacts  and 
(c) ,  consumed  in  the  electrolyte  itself. 

(3)  If  interest  on   capital  is   6  per  cent,   what  expenditure 
would  be  justified  on  means  for    (a),  reducing  the  conductor 
losses   (b),  reducing  the  contact  losses   (c),  reducing  the  elec- 
trolyte losses. 

(4)  If  the  ampere  efficiency  of  deposition  could  be  increased 
5  per  cent  by  reducing  the  current  density  20  per  cent,  would  it 
pay? 

Solution : 

(1)  Copper  deposited  in  1  tank  per  day 

3800X0.82  =  3118  oz.  =        195  Ibs. 

In  200  tanks 

200X195  =  39,000    " 
Power  used 

3800X67+1000  =       255  kw. 

Cost  of  power  per  day 

255  X  $157  +-365  =  $109.70 
Cost  of  power  per  2,000  Ibs.  copper 

$109.70  +  19.5=        $5.63  (1) 

(2)  The  voltage  lost  in  the  conductors  is 

67  -  200(0.0368  +  0.2567)  =  67  -  58.7  =  8.3  volts 

and  the  power  thus  lost 

3800X8.3+1000    =         31.54  kw. 
Cost  per  year       31.54X  $157    =  $4,952.00 
Cost  per  day         $4952+    365    =        $13.57 
Cost  per  ton  of  copper  produced  =          $0.70  (2a) 

Power  lost  in  the  contacts 

200  X  0 . 0368  X  3800  + 1000  =         27.97  kw. 
Cost,  per  year  =  $4,391.00 
Cost,  per  day    =        $12.03 
Cost,  per  ton     =         $0.62  (26) 

Power  consumed  in  the  electrolyte 

200  X  0 . 2567  X  3800  +  1000  =  195  kw. 

Cost,  per  year  =  $30,615.00 
Cost,  per  day  =  $83.88 
Cost,  per  ton  =  $4.30  (36) 


THE  METALLURGY  OF  COPPER.  565 

(3)  The  cost  of  the  power  lost  per  year,  divided  by  0.06, 
will  give  the  capital  investment  representing  that  yearly  outlay. 

This  would  be 

Cost  of  Power      Capitalized. 

Conductor  resistances $4,952  $82,530 

Contact  resistances 4,391  73,180 

Electrolyte  resistances 30,615  510,250 

In  the  specific  cases  in  point  each  1  per  cent  of  the  power 
consumption  which  could  be  saved  by  improvements  of  a  per- 
manent character,  not  subject  to  depreciation,  would*  justify 
capital  expenditures  in  the  plant  up  to  the  following  amounts: 

Conductor  .resistances -. $825 

Contact  resistances 732 

Electrolyte  resistances 5,100 

If  the  improvements  were  liable  to  depreciation  in  use,  such 
as  larger  tanks,  contact  clamps,  etc.,  the  depreciation  must  be 
added  to  the  inte'rest  to  find  the  justifiable  capital  expenditure. 
Assuming  depreciation  10  and  20  per  cent,  respectively,  this 
plus  interest  amounts  to  16  and  26  per  cent,  and  the  justified 
expenditures  must  be  less  than  the  following  amounts  for  1  per 
cent  saving  of  power  in  each  direction: 

10%  20% 

Depreciation.  Depreciation. 

Conductor  resistances $309 .50  $190 . 50 

Contact  resistances 274 . 50  169 . 00 

Electrolyte  resistances 1,913 . 50  736 . 00 

There  is  here  a  wide  field  for  experiment  and  improvement. 
In  the  case  in  point,  making  all  allowances,  one  would  be  justi- 
fied in  expending  in  permanent  plant  for  each  tank  at  least 
$1.00  for  every  1  per  cent  reduction  which  could  be  made  in 
the  power  loss  in  the  conductors,  $0.85  for  every  1  per  cent 
gained  in  reducing  contact  resistances  and  $3.68  for  each  1 
per  cent  reduction  in  the  resistance  of  the  electrolyte.  To  be 
more  specific,  assume  the  main  conductors  to  be  proportioned 
to  carry  2,000  amperes  per  square  inch  of  cross-section.  Their 
resistance  in  the  case  in  point  is  8. 3  -5-  3, 800  =  0.0022  ohm, 
cross-section  1.9  square  inches,  resistance  per  running  inch 
0.00000067 -f- 1.9  =  0.00000035  ohm,  and  total  length  0.0022-v- 
0.00000035  =  6,286  inches  =  524  feet. 


566  METALLURGICAL  CALCULATIONS. 

The  weight,  at  62.5  X  8.9 -v- 1,728  =  0.322  Ibs.  per  cubic  inch  is 

6,286X1.9X0.322  =  3,846  Ibs. 
which  at  $0.20  per  Ib.  costs 

3,846X0.20  =  $769.20 

Since  there  is  practically  no  depreciation  (except  change  in 
market  value)  in  these  copper  conductors,  if  we  increased  the 
cross-section  of  the  conductors  100,  200,  300,  and  400  per  cent, 
respectively,  we  would  decrease  their  resistance  50,  67,  75  and 
80  per  cent,  respectively,  and  arrive  at  the  following  figures:' 

Per  Ton  of  Copper  Produced. 


Per  Cent 

\ 

Interest 

Per  Cent 

Decreased 

on 

Increased 

Resistance 

Increased 

Increased 

Decreased 

Size  of 

of 

•Cost  of 

Cost  of 

Cost  of 

Conductors. 

Conductors.  Plant. 

Plant. 

Power. 

Gain. 

100 

50 

$769 

$0.0065 

$0.35 

$0.35 

200 

67 

1,538 

0.013 

0.47 

0.46 

300 

75 

2,307 

0.019 

0.52 

0.50 

400 

80 

3,077 

0.026 

0.56 

0.53 

900 

90 

6,923 

0.058 

0.63 

0.57 

1,900 

95 

14,613 

0.123 

0.67 

0.54 

We  thus  see  that  for  this  plant,  with  power  costing  $157  a 
kilowatt-year  and  copper  bar  $0.20  per  pound,  a  great  reduc- 
tion in  the  power  costs  up  to  $0.57  per  ton  of  copper  produced 
can  be  made  by  making  the  main  conductors  ten  times  as  large ; 
that  is,  by  making  them  carry  only  200  amperes  per  square 
inch  of  section  instead  of  2,000. 

At  a  locality  like  Great  Falls,  Montana,  where  the  water 
power  does  not  cost  over  one-tenth  the  above  price,  let  us  say 
$15.70  per  kilowatt-year,  the  decreased  cost  of  power,  in  the 
above  table  would  be  one-tenth  the  above  figures — there  would 
be  only  $0.07  per  ton  of  copper  to  save  if  all  were  saved,  and 
it  can  easily  be  seen  that  the  gains  would  be 

for  100  per  cent  increased  size $0.029 

"200         "                "           "    0.034 

"300         "                "           "    0.033 

«    400         "                "           "  0.030 


THE  METALLURGY  OF  COPPER.  567 

We  reach  the  maximum  saving  in  this  case  by  increasing  the 
size  200  per  cent,  giving  a  current  density  of  660  amperes  per 
square  inch  of  conductor  section. 

It  can  easily  be  seen  that  the  cost  of  copper  bars,  rate  of  in- 
terest, and  cost  of  power  are  all  three  factors  in  determining 
the  economic  size  to  be  given  the  conductors.  In  general 
they  are  made  much  smaller  than  they  should  be  because  of 
lack  of  capital  to  invest  in  them.  As  soon  as  a  refinery  is  run- 
ning and  paying  dividends  part  of  its  profits  should  be  ap- 
plied to  permanent  investment  in  heavier  copper  conductors  in 
order  to  reduce  the  running  costs. 

(4)  If  the  ampere  efficiency  were  increased  5  per  cent  and 
the  current  density  20  per  cent,  all  other  data  being  constant 
except  those  affected  by  these  changes,  the  output  of  copper 
would  be  87  per  cent  of  the  theoretical  output  of  3,040  amperes 
instead  of  82  per  cent  of  the  theoretical  output  of  3,800  am- 
peres. The  output  of  copper  per  day  would  therefore  be,  in 
place  of  39,000  pounds: 

3,040X0.87X200-;- 16  =  33,060  Ibs.  per  day. 

The  power  cost  will  be  reduced  20  per  cent  by  the  lower 
amperage  used,  that  is,  be  $5.63X0.80  =  $4.50  per  ton — a  sav- 
ing of  $1.13.  Whether  it  will  pay  to  do  this  depends  of  what 
the  other  fixed  charges  of  the  plant  are.  Decreasing  the  out- 
put 15  per  cent  increases  the  fixed  charges  per  ton  of  copper 
produced  12  per  cent.  It  is  then  a  question  whether  12  per 
cent  of  the  fixed  charges  in  the  first  case  is  greater  or  less  than 
$1.13.  If  the  fixed  charges  amounted  to,  say,  a  usual  figure 
of  $5.00  per  ton,  decreasing  the  output  15  per  cent  would  in- 
crease this  charge  per  ton  of  copper  produced  12  per  cent, 
or  $0.60;  but  if  there  is  a  power  saving  of  $1.13  there  is  a  mar- 
gin of  $0.53  per  ton  in  favor  of  making  the  change.  If  the 
power  cost  were  only  $0.56  per  ton  of  copper  it  would  just 
barely  pay  to  make  the  change.  Everything  depends  upon  the 
relative  cost  of  power  and  of  the  other  fixed  charges. 

Energy  Lost  in  Conductors. 

We  have  already  touched  on  this  subject,  but  it  deserves 
still  greater  attention.  The  resistivity  of  copper  is  0.00000167 
ohm  per  centimeter  cube,  or  0.00000067  ohm  per  inch  cube. 


568  METALLURGICAL  CALCULATIONS. 

Given  a  conductor  of  certain  length  its  resistance  in  ohms  is 
0.00000167  multiplied  by  its  length  in  centimeters  and  divided 
by  its  cross  sectional  area  in  square  centimeters.  The  similar 
calculation  can  be  made,  for  the  inch  units.  The  drop  in  po- 
tential in  the  bus-bars  is  their  resistance  multiplied  by  the 
amperes  passing,  and  the  power  consumption  is  the  voltage 
drop  into  the  amperes  passing  or  the  resistance  in  ohms  into  the 
square  of  the  amperes  passing.  The  watts  thus  expended 
may  be  expressed  in  kilowatts,  or  transposed  into  horse-power 
by  dividing  by  746.  The  heat  generated  in  the  conductors  is, 
per  second,  in  gram  calories,  the  number  of  watts  expended  in 
them  multiplied  by  0.2389.  The  rate  of  rise  of  temperature  of 
the  conductors,  leaving  out  radiation  and  conduction  to  the 
air,  will  be,  per  second,  the  gram  calories  generated  in  the  con- 
ductors divided  by  the  heat  capacity  of  the  conductors  per  de- 
gree, which  in  the  case  of  copper  is  volume  in  cubic  centi- 
meters into  0.837.  This  rise  in  temperature  will  continue  until 
by  radiation  and  conduction  to  the  air  the  conductors  lose  as 
much  heat  per  second  as  the  current  generates  in  them.  This 
will  vary  with  the  shape  of  the  cross  section  of  the  conductors, 
because  the  loss  of  heat  is  proportional  to  the  surface  exposed, 
and  the  area  of  exposed  surface  for  a  given  cross  section  is 
greater  the  thinner  the  conductor  and  is  a  minimum  for  a 
round  conductor.  The  amount  of  heat  thus  lost  can  be  cal- 
culated by  the  principles  explained  in  Metallurgical  Calcula- 
tions, Part  I.,  pp.  172-186.  If  it  is  desired  to  get  along  with 
the  minimum  heating  of  the  conductors,  they  should  be  as  flat 
and  thin  as  possible;  if  heating  can  be  ignored  they  may  be  of 
the  cheap  round  shape.  It  should  not  be  forgotten  also  that 
the  resistivity  of  copper  increases  as  it  gets  hotter  in  direct 
proportion  to  its  absolute  temperature  (C°  +  273),  so  that 
the  heating  effect  really  aggravates  itself  from  this  increased 
resistance. 

Problem  121. 

A  copper  conductor  is  desired  to  carry  4,000  amperes,  300 
meters.  Round  bars  sell  at  18  cents  per  pound.  Interest  rate 
10  per  cent.  Cost  of  electric  power  as  delivered  $50  per  kilo- 
watt year.  Average  temperature  of  surrounding  air  20°  C. ; 
air  still.  Conductivity  of  the  copper  600,000  reciprocal  ohms 
per  centimeter  cube.  Specific  gravity  of  copper  8.9,  specific 
heat  0.094. 


THE  METALLURGY  OF  COPPER.  569 

Aluminium  conductors  to  replace  same  have  conductivity  of 
375,000  reciprocal  ohms,  specific  gravity  2.60,  specific  heat 
0.230;  cost  of  round  bars  36  cents  per  pound. 

Required : 

(1)  The    cross-sectional    area    of   the    conductors    of    copper 
which  will  give  the  minimum  cost  of  power  and  interest  on  the 
investment. 

(2)  The  same  for  aluminium. 

(3)  The   running  temperature   of  the   conductors   of   copper 
above  the  room  temperature. 

(4)  The  same  for  aluminium. 
Solution : 

(1)   Let  S  =  cross  section  of  conductor,  in  sq.  cm. 
Resistance  of  copper  conductor 

0. 00000167 X  (300 X  100) -S  =  0.0501  oh^ 

o 

Power  expended  in  the  conductor: 

0.0501  801,600 

— ^ — X  (4,000) 2  = ^ watts. 

o  o 

Cost  of  power  thus  expended,  per  year 

801,600  40,080  A 

-r-  1,000X50  =  — ^ —  dollars. 


S 
Weight  of  copper  in  conductor: 

(300  X  100)  X  S  X  8 . 9  -*- 1 ,000  =  267  X  S  kilograms. 
Cost  of  conductor,  if  round 

267.  SX  0.18X2.205  =  106  XS  dollars. 

Interest  on  cost  of  conductor,  at  10%       =  10.6XS  dollars. 
Sum  of  cost  of  power  and  interest  on  investment: 

4°'080+(10.6XS) 


S 

Solving  for  S  a  minimum,  we  find  S  =  61.5  sq.  cm.  (1) 

Total  costs,  substituting  S  above 

$652  (for  power)  +$652  (interest)  =  $1,304  per  year. 


570  METALLURGICAL  CALCULATIONS. 

(2)   Resistance  of  aluminium  conductor 

0.  00000267  X  (300X100)^-8  =  °^801  ohm. 

o 

Power  expended  in  the  conductor 

0.0801  ^x,  nnnv,       1,281,600 

—  g  —  X  (4,000)2  =  —  —  g  -  watts. 

Cost  of  power  thus  expended,  per  year 

1,281,600  64,080  , 

-  ~  -  -^1,000X50  ==  —  ^  —  •  dollars. 
o  o 

Weight  of  aluminium  in  conductor 

(300  X  100)  X  8x2.64-1,000  =  78x8.  kilograms. 
Cost  of  conductor,  if  round 

78.  8.  X  0.36X2.205  =  62x8.  dollars. 

Interest  on  cost  of  conductor,  at  10%       =  6.2x8.  dollars. 
Sum  of  cost  of  power  and  interest  on  investment: 


Solving  for  S  =  a  minimum:  S  =  101.2  sq.  cm.         (2) 

Total  costs,  substituting  S  above 

$633  (for  power)  +$627  (interest)  =  $1,260  per  year, 
(3)  Power  expended  in  copper  conductor 

=  13,035  watts. 


bl.o 
Heat  generated  in  conductor,  per  second 

13,035X0.2389  =  3,114  gm.  cal. 
Area  of  conductor  surface 

(300  X  100)  X  3.  14  (x/61.5  4-  0.7854)  = 
=  30,000X27.8  (cm.  circumference)  =  834,000  sq.  cm. 


THE  METALLURGY  OF  COPPER.  571 

Loss  of  heat  per  second  by  contact  with  still  air,  per  1°  difference 

0.000056X834,000  =  46  .  7  calories. 
Rise  of  temperature,  excluding  radiation  loss 

3,114*46.7  =  67°. 

The  radiation  loss  must,  however,  be  considered.  From 
copper  it  is  0.00068  gram  calories  per  second  per  sq.  centi- 
meter, if  the  surroundings  are  at  0°  and  the  hot  body  at  100°, 
this  is  approximately  0.0000068  gram  calories  per  each  degree 
in  this  range.  We  have  the  combined  conduction  and  radia- 
tion losses  per  1°  per  second 

0  .  000056  +  0  .  000007  =  0  .  000063  calories       • 
and  the  rise  in  temperature  of  the  conductor 

3,1  14  -KG.  000063X834,000)  =  59°. 
The  conductors  would  therefore  run  at  a  temperature  of 

20  +  59  =  79°    C.  (3) 

=  174°  F. 

A  correction  of  the  second  order  would  be  to  take  into  con- 
sideration the  fact  that  at  79°  the  copper  would  have  less  con- 
ductivity than  at  20°,  its  resistivity  at  79°  being 


0.  00000167  X  =  0.0000020  ohm. 


This  would  make  the  required  section  for  minimum  expense 
68  sq.  cm.  instead  of  61.5,  and  would,  by  increasing  the  sur- 
face, make  a  different  rise  in  temperature.  The  power  ex- 
pended would  be  the  equivalent  of  3,373  gram  calories  per 
second,  and  the  revised  temperature  of  the  conductor  would  be 

(3,373  -r-52.  54)  +20  =  64  +  20  =  84°  C. 
(4)  Power  expended  in  the  aluminium  conductor 

=12,665  watts. 


Heat  generated  in  conductor,  per  second 

12,665  X  0  .  2389      =  3,026  calories. 


572  METALLURGICAL  CALCULATIONS. 

Area  of  conductor  in  square  centimeters         =  1,074,000  sq.  cm. 
Conduction  and  radiation  losses,  per  1° 

(0.000056  +  0. 000010)  X  1,074,000       =70.88  grm.  cal. 
Rise  in  temperature  of  conductor 

3,026-^70.88      =  43°. 
Working  temperature  of  conductor 

43  +  20    =  63°  C. 

The  aluminium  conductor  thus  is  seen  to  work  cooler  than 
the  copper,  and  to  admit  of  a  lower  minimum  of  working  costs. 

v 

Electric  Smelting  of  Copper  Ores. 

The  use  of  electrothermal  processes  for  smelting  copper  ores 
has  been  proposed  in  recent  years,  and  in  a  few  cases  experi- 
mentally tested.  The  whole  question  depends  on  the  recog- 
nition and  appreciation  of  a  few  fundamental  facts.  By  far 
the  greater  bulk  of  all  copper  ores  are  sulphide  ores,  and  in  many 
cases  the  sulphur  and  iron  present  are  sufficient  heat  producers 
to  allow  of  the  smelting  of  the  ore  simply  by  the  heat  of  their 
oxidation,  if  the  operation  is  skillfully  conducted.  Such  is  the 
basic  principle  of  pyritic  smelting,  and  whenever  it  can  be 
applied  it  is  very  economical  to  do  so,  and  electric  processes 
have  no  chance  of  an  application. 

Other  ores  of  copper  contain  so  small  an  amount  of  the 
metal  that  the  result  of  the  smelting  down  of  the  whole  to  a 
fluid  mass,  by  any  method  of  fusion,  would  not  repay  the  cost 
of  the  operation.  Some  such  ores  can  be  treated  by  aqueous 
and  other  methods  not  involving  fusion,  with  a  margin  of 
profit  to  pay  for  the  operation. 

The  only  field  for  electrothermic  processes  appears  to  be  in 
the  smelting  down  of  ores  carrying  too  little  sulphur  to  be 
"  pyritically  "  smelted,  and  which  require,  as  usually  treated, 
the  use  of  carbonaceous  fuel  to  assist  the  fusion.  Wherever  such 
fuel  is  expensive,  and  electric  power  may  be  obtained  at  a  low 
price,  an  electrothermic  process  might  be  theoretically  possible 
and  profitable. 

Such  conditions  may  easily  occur  in  the  vicinity  of  copper 
mines.  Situated  frequently  among  the  mountains,  remote  from 


THE  METALLURGY  OF  COPPER.  573 

railways  and  cheap  fuel,  they  frequently  are  near  large  water 
powers  which  would  furnish  cheap  electric  power.  Some  may 
be  even  so  situated  that  concentration  of  a  very  lean  sulphide 
ore  to  matte  by  use  of  fuel,  or  by  mechanical  concentration, 
would  be  unprofitable,  and  yet  a  concentration  or  simple  melt- 
ing to  matte  by  electrically-generated  heat  be  profitable. 

There  are,  so  far  as  the  writer  knows,  no  electrothermal 
processes  yet  in  commercial  operation  on  copper  ores,  yet  they 
have  been  tested  experimentally  with  promising  results.  Vat- 
tier,  for  instance,  conducted  experiments  at  the  Li  vet  works  of 
the  Compagnie  Electrothermique  Keller  et  Leleux,  on  April  23, 
1903,  in  the  presence  of  several  distinguished  metallurgists, 
such  as  Mr.  Stead,  of  Middlesborough ;  Mr.  Allen,  of  Sheffield, 
and  Mr.  Saladin,  of  the  Creusot  works.  The  ores  were  Chilian 
ores  sent  by  the  Chilian  Government  to  test  the  possibility  of 
electric  smelting,  taken  from  the  "  Vulcan  "  mine,  owned  by  G. 
Denoso,  and  low-grade  ore  from  Santiago.  At  the  mines,  coke 
costs  $20  per  ton,  but  the  slopes  of  the  Andes  afford  a  fine 
opportunity  for  developing  large  water  powers  cheaply;  it  is 
estimated  that  the  total  cost  of  electric  power  delivered  should 
not  exceed  $6  per  kilowatt-year.  A  detailed  account  of  these 
tests  can  be  found  in  the  appendix  to  the  Report  of  the  Can- 
adian Commission  on  Electric  Smelting,  1904. 

Problem  122. 

Vattier  mixed  his  rich  and  poor  Chilean  ores  to  a  charge 
consisting  of 

Per  Cent. 

Copper 5. 10 

Sulphur. ...... 4. 13 

Iron : 28.50 

Manganese .7 . 64 

Silica 23.70 

Alumina 4 . 00 

Carbonic  acid 4.31 

Lime 7.30 

Magnesia 0.33 

Phosphorus 0.05 

This  was  charged  directly  into  a  shaft  furnace  1.8  meters 
long,  0.9  meter  wide  and  0.9  meter  high.  The  melted  material 


574  METALLURGICAL  CALCULATIONS. 

ran  out  into  a  forehearth  1.2  meters  long,  0.6  meter  wide  and 
0.6  meter  high.  Two  carbon  electrodes,  30  centimeters  square 
by  170  centimeters  long,  were  used  in  the  smelting  chamber, 
and  two  electrodes,  25  centimeters  square  by  100  centimeters 
long,  were  put  in  the  forehearth  to  reheat  it  for  tapping. 
The  matte  obtained  analyzed: 

Per  Cent. 

Silica 0.80 

Alumina 0 . 50 

Iron 24.30 

Manganese 1 . 40 

Sulphur 22.96 

Copper 47.90 

The  slag  contained : 

Silica ..27.20 

Alumina 5 . 20 

Lime 9.90 

Magnesia 0.39 

Iron 32.50 

Manganese 8.23 

Sulphur 0.57 

Phosphorus 0.06 

Copper 0.10 

The  current  used  was  4,750  amperes  at  119  volts;  power 
factor  =  0.9;  8,000  kilograms  of  ore  mixture  were  smelted  in 
8  hours;  consumption  of  electrodes  50  kilograms. 

Required : 

(1)  A  balance  sheet  of  materials  entering  and  leaving  the 
furnace  per  hour. 

(2)  A  balance  sheet  of  the  heat  development  and  distribution 
in  the  furnace. 

(3)  The  saving  in  cost  of  treatment  per  ton  of  ore,  assuming 
electrodes  to  cost  4  cents  per  kilogram,  coke  $20  per  metric 
ton,  and  that  when  coke  is  used  one-third  of  its  calorific  power 
leaves  the  furnace  in  the  escaping  hot  gases.     Assume  all  other 
costs  similar,  except  that  air  blast  costs  $0.10  per  ton  of  ore 
smelted  additional.     Electric  power  $6  per  kilowatt-year. 

(4)  If  the  resulting  slag  were  smelted  electrically  for  ferro 


THE  METALLURGY  OF  COPPER.  575 

silicon,  at  an  expenditure  of  0.75  electric  horse  power-year  per 
ton  of  ferro-silicon  obtained,  and  other  smelting  costs  were 
$10  per  ton,  how  much  ferro-silicon  would  be  obtained  per  ton 
of  copper  ore  treated,  and  what  would  be  its  cost? 

Solution : 

(1)  Balance  sheet  per  1,000  kg.  of  ore. 

Charges              Matte           Slag  Gases 
Ore:  1,000  kg. 

Silica 237   kg.  1  236 

Alumina 40     "  0.5  39.5  

Iron 285     "  25  260  

Manganese 76     "  1.5  74.5  

Sulphur 41      "               24                   5  ,       12 

Copper 51     "               50                   1  .... 

Carbonic  acid 43     "                43 

Lime 73     "  73  

Magnesia 3     "  3  

Oxygen '. 137     "                2                 74  61 

Electrode : 

Carbon..                       6     "  6 


104  766  122 

Composition  of  Slag. 

Calculated.  Analyzed. 

Silica 29.5  per  cent.  27.2  per  cent. 

Alumina 5.0        "  5.2 

Lime 9.2        "  9.9 

Iron 32.5        "  32.5 

Manganese 9.3        "  8.2 

Sulphur 0.6        "  0.6 

Heat  Development. 
Electric  energy  per  hour: 

4,750  X  1 19  X  0 . 9  =          509  kilowatt-hours. 

509X860  =  437,750  Cal.  per  hour. 

Oxidation  of  carbon: 

6X4,300  =    25,800     "      "       a 


463,550    a 


576  METALLURGICAL  CALCULATIONS. 

Heat  Distribution. 
Heat  in  matte : 

104X270  =     28,080  Calories. 

Heat  in  slag: 

796X400  =  318,400 

Heat  in  gases        =     30,000         " 
Loss  by  radiation 

and  conduction  =     87,070         " 


463,550 

The  radiation  and  conduction  loss  is  19  per  cent  of  the  total 
heat  generated  in  the  furnace — a  satisfactory  showing.  The 
useful  effect  of  the  furnace  is  nearly  75  per  cent,  reckoning  as 
usefully  applied  heat  that  contained  as  sensible  heat  in  the 
melted  matte  and  slag.  Nearly  90  per  cent  of  this  usefully 
applied  heat  is  contained  in  the  slag. 

(3)  If  coke  were  used  instead  of  electric  heat,  enough  coke 
would  have  to  be  used  to  furnish  433,550  Calories  plus  the  heat 
in  the  gases.  Since  the  latter  is  assumed  to  be  one-third  of  its 
calorific  power,  the  coke  must  have  a  calorific  power  of  650,325 
Calories,  which  at  7,000  Calories  per  kilogram  would  require 
93  kg.  of  coke  per  1,000  of  ore  smelted.  The  costs  of  coke  and 
blowing  engine  would  therefore  be 

Coke,  93X0.02  =  $1.96 

Engine,  0.10 


Sum  $2.06 

The    cost    of   electrodes    and    electric    energy,    which   would 
replace  coke  and  blowing  engine,  would  be: 

Electrodes,         6  kg.  X  $0.04  =$0.24 
Power  509kw.X   A=    0.35 


Sum,       $0.59 
Saving,       $1.47 

(4)  The  resulting  slag  contains,  according  to  the  balance 
sheet,  260  kg.  of  iron,  74.5  kg.  of  manganese,  and  236  kg.  of 
silica  equal  to  126  kg.  of  silicon.  It  should  produce,  if  com- 


THE  METALLURGY  OF  COPPER.  577 

pletely  reduced,  460  kg.  of  alloy  (omitting  the  carbon  it  might 
contain)  of  the  approximate  composition: 

Fe  260  kg.  =  56  per  cent. 

Mn  74  kg.  =  13 

Si  126  kg.  =  28         a 

460  kg. 

Cost  of  power  $6.00  Xj  =  $4.50  per  ton. 
Other  costs  10.00       " 


Total  costs  per  ton  $14 . 50 

This  cost  is  very  low,  because  of  the  very  low  figure  assumed 
for  power  cost.  However,  the  whole  calculation  points  to 
the'  possibility  of  great  saving  in  some  localities  in  smelting 
down  copper  ores  by  electrically  applied  heat,  and  the  possi- 
bility of  cheaply  producing,  at  the  same  spot,  valuable  ferro- 
silicon  as  a  by-product. 


CHAPTER  II. 
THE  METALLURGY  OF  LEAD. 

The  extracting  of  lead  from  its  ores  and  the  refining  of  the 
crude  metal  to  commercial  lead,  constitutes  the  metallurgy  of 
lead.  The  chief  ore  is  galena,  lead  sulphide,  PbS;  but  the 
carbonate,  cerussite,  PbCO2,  and  the  sulphate,  Anglesite,  PbSO4, 
occur  at  some  places  in  important  quantities;  the  silicate,  phos- 
phate, molybdate,  tungstate,  chloride  and  native  lead  are  rare 
minerals. 

The  reduction  of  the  oxidized  lead  ores,  such  as  often  occur 
at  the  outcrops  of  sulphide  veins,  is  very  simple;  carbon,  the 
cheapest  and  most  universal  reducing  agent,  reduces  them 
satisfactorily  at  a  red  heat,  with  some  loss  of  lead  by  volatiliza- 
tion. The  sulphide  of  lead,  however,  is  not  reducible  by  carbon; 
it  requires  other  treatment.  If  we  tabulate  the  most  common 
sulphides  according  to  their  thermochemical  heats  of  formation, 
expressed  per  unit  weight  of  sulphur  held  in  combination  (32 
kilograms),  we  find  the  common  metals  arranged  as  follows: 

Calories. 

Potassium  (K2,  S) 103,500 

Calcium  (Ca,  S) 94,300 

Sodium  (Na2,  S) 89,300 

Manganese  (Mn,  S) 45,600 

Zinc  (Zn,  S) 43,000 

Cadmium  (Cd,  S) 34,400 

Iron  (Fe,  S) 24,000 

Cobalt  (Co,  S) 21,900 

Copper  (Cu2,  S) 20,300 

Lead  (Pb,  S) 20,200 

Nickel  (Ni,  S) 19,500 

Mercury  (Hg,  S) 10,600 

Hydrogen  (H2,  S) 4,800 

Silver  (Ag2,  S) 3,000 

578 


THE  METALLURGY  OF  LEAD.  .  579 

A  glance  at  this  table  shows  us  that  some  metals  unite  with 
sulphur  more  energetically  than  lead  does,  and  others  less 
energetically,  and  points  to  the  theoretical  possibility  of  de- 
composing lead  sulphide  by  the  agency  of  copper,  cobalt,  iron, 
cadmium,  zinc,  etc.  Of  these  agents,  iron  is,  of  course,  the 
cheapest  and  most  available,  and  can  be  used  to  reduce  lead 
sulphide  with  formation  of  iron  sulphide. 

PbS  +  Fe  =  FeS  +  Pb 
-  20,200       +  24,000 

Showing  an  excess  heat  development  per  207  parts  of  lead 
set  free,  of  24,000-20,200  =  3,800  Calories.  If  lead  sulphide 
and  iron  are  brought  together  at  a  red  heat,  at  which  heat  the 
sulphide  is  molten,  they  react  energetically  with  evolution  of 
heat — enough  theoretically  to  raise  the  temperature  of  the 
207  kg.  of  lead  and  88  kg.  of  iron  sulphide  some  280°  C.  The 
reaction  is  fairly  complete  if  sufficient  reducing  agent  is  present 
and  time  is  allowed,  so  that  it  can  be  used  in  assaying  to  deter- 
mine lead  by  fire  assay,  but  in  commercial  practice  more  or  less 
undecomposed  lead  sulphide  always  remains  in  the  iron  sul- 
phide, forming  a  sort  of  double  sulphide  or  iron-lead  matte. 

It  is  interesting  to  note,  en  passant,  that  lead  vigorously 
reduces  silver  sulphide,  the  liberated  silver  alloying  with  the 
excess  of  lead.  This  reaction  is  accompanied  by  a  large  evolu- 
tion of  heat,  as  the  table  shows,  is  the  basis  of  the  ordinary 
assaying  methods  for  silver  ores,  and  is  used  commercially  to 
extract  silver  from  its  ores  wherever  lead 'is  plentiful. 

The  affinity  of  lead  for  oxygen  is  a  no  less  interesting  sub- 
ject, since  it  concerns  not  only  the  reduction  of  oxide  ores 
but  also  the  roasting  of  sulphide  ores  to  oxide,  and  very  largely 
controls  the  refining  of  impurities  from  lead  by  methods  in- 
volving oxidation.  The  heat  of  combination  of  some  of  the 
more  common  elements  with  oxygen,  expressed  per  unit  weight 
of  16  kilograms  of  oxygen  held  in  combination,  is  as  follows: 

Calories. 

Magnesium  (Mg,  O) 143,400 

Calcium  (Ca,  O) 131,500 

Aluminium  J  (AP,  O3) . 130,870 

Sodium  (Na2,  O) 100,900 

Silicon  i  (Si,  O2) 98,000 


580  •;"     METALLURGICAL  CALCULATIONS. 

Calories. 

Manganese  (Mn,  O) 90,900 

Zinc  (Zn,  O) 84,800 

Tin  }(Sn,  O2) 70,650 

Iron  (Fe,  O) 65,700 

Iron  J(Fe2,  O3) 65,200 

Nickel  (Ni,  O) 61,500 

Hydrogen  (H2,  O) 58,060 

Antimony  J(Sb2,  O3) 55,630 

Arsenic  J(As2,  O3) 52,130 

Lead  (Pb,  O) 50,800 

Carbon  |(C,  O2) 48,600 

Bismuth  J(Bi2,  O3) 46,400 

Copper  (Cu2,  O) 43,800. 

Sulphur  }(S,  O2) 34,630 

Sulphur  i(S,  O3) 30,630 

Carbon  (C,  O) 29,160 

Mercury  (Hg,  O) 21,500 

Silver  (Ag2,  O) 7,000 

Reduction  of  Lead  Oxide. — An  inspection  of  above  table 
shows  that  lead  oxide  is  a  weak  oxide,  weaker  than  the  oxides 
of  most  of  the  common  metals.  It  is  reduced  to  metallic  lead 
by  many  reagents  with  evolution  of  heat.  It  is  also  reduced 
by  some  weaker  reagents  with  absorption  of  heat,  provided 
that  the  oxide  formed  is  gaseous  and  the  necessary  heat  energy 
is  supplied  from  outside.  For  instance, 

PbO  +  C  =  Pb  +  CO 
-50,800  +29,160 

involves  an  absorption  of  21,640  Calories,  or  several  times  as 
much  heat  as  is  necessary  to  raise  the  reacting  substances, 
PbO  and  C,  to  the  reacting  temperature,  a  red  heat.  The 
reaction  is  endothermic,  absorbing  heat,  and  therefore  only 
progresses  in  measure  as  the  necessary  calories  are  supplied 
from  outside.  The  fact  that  all  the  substances  concerned  are 
liquid  or  solid  except  the  product  CO,  gives  a  predisposing 
cause  which  facilitates  the  reaction,  that  is,  the  reaction  can 
only  go  one  way,  since  the  CO  gas  escapes  from  the  sphere  of 
action  as  soon  as  formed. 


THE  METALLURGY  OF  LEAD.  581 

Oxidation  of  Lead  Sulphide. — When  PbS  is  roasted,  that  is, 
heated  to  redness  with  free  access  of  air,  both  the  sulphur  and 
the  lead  tend  to  oxidize,  generating  a  large  heat  of  oxidation, 
against  which  there  is  absorbed  only  the  comparatively  feeble 
heat  of  decomposition  of  PbS.  The  equation  may  be  discussed 
as 

2PbS  +  3O2  =  2PbO  +  2SO2 

-  (20,200)  +  2  (50,800) +2  (69,260) 

The  net  heat  evolved  in  the  equation  is  240,120  —  40,400  = 
199,720  Calories,  which  is  33,290  Calories  per  unit  weight  (16 
kg.)  of  oxygen  used,  or  418  Calories  for  each  kilogram  of  lead 
sulphide  oxidized.  This  great  heat  of  oxidation,  evolved  in 
roasting,  is  quite  sufficient,  in  fact  more  than  sufficient,  to 
provide  the  heat  required  for  self-roasting  without  the  use  of 
any  other  fuel,  the  chief  difficulty  is  really  to  keep  the  charge 
from  getting  too  hot  and  melting  everything  down  to  a  liquid 
before  the  roasting  is  anywhere  near  complete.  In  the  ordinary 
hand-worked  reverberatory  roaster,  the  oxidation  is  so  slow 
that  the  fire  on  the  grate  really  controls  the  temperature  on  the 
hearth,  and  the  temperature  of  the  roasting  ore  can  be  regu- 
lated accordingly.  Where  the  roasting  is  done  quickly,  by  an 
air  blast,  as  in  "  Pot  Roasting,"  the  temperature  is  kept  down 
somewhat  by  previously  roasting  off  part  of  the  sulphur,  or  by 
liberally  wetting  the  charge,  or  by  using  limestone  in  with  the 
ore,  to  absorb  by  its  decomposition  into  CaO  and  CO2  a  large 
part  of  the  excess  heat — while  the  melting  down  of  the  charge 
is  prevented  by  the  mechanical  interference  presented  by  inter- 
mixed, inert  and  infusible  material,  such  as  silica,  lime,  etc., 
which  simply  prevents  the  really  melted  globules  of  sulphide  from 
running  together  and  thus  melting  down  to  a  liquid  mass. 

An  interesting  variation  is  the  roasting  of  lead  sulphide  to 
sulphate;  some  of  this  always  forms,  due  to  the  high  formation 
heat  and  difficult  decomposability  of  the  sulphate. 

PbS  +  2O2=PbSO< 
-  (20,200)     +  (215,700) 

The  net  heat  evolved  is  215,700-20,200  =  195,500  Calories, 
or  48,875  Calories  per  unit  weight  (16  kg.)  of  oxygen  used,  or 
818  Calories  for  every  kilogram  of  lead  sulphide  so  oxidized. 


582  METALLURGICAL  CALCULATIONS. 

This  is  a  high  heat  of  oxidation,  and  explains  the  great  ten- 
dency to  form  sulphate  observed  during  roasting.  If  the  con- 
ditions could  be  found  whereby  only  this  reaction  occurred,  the 
sulphate  roasting  could  be  easily  made  automatic  without  out- 
side fuel.  There  is  needed,  at  the  present  time,  a  careful  lab- 
oratory investigation  of  the  conditions,  mechanical,  physical, 
chemical  and  thermal,  for  the  roast  exclusively  to  sulphate — 
just  as  a  bit  of  badly  needed  metallurgical  information. 

Double  Reactions. — A  large  part  in  the  metallurgy  of  lead  is 
played  by  double  reactions,  such  as  the  following: 

PbS  +  PbSO4  =  2Pb  +  2SO2 
PbS  +  2PbO   =  3Pb  +  SO2 
PbS  +  3PbSO4  =  4PbO  +  4SO2 

On  roasting  lead  sulphide  for  a  short  time,  either  lead  or 
lead  sulphate  or  mixtures  of  these  are  formed,  according  to  the 
temperature  and  excess  of  air  provided.  With  large  excess  of 
air  and  low  temperature,  and  especially  in  presence  of  infusible 
materials  which  act  as  catalyzers  (i.e.,  which  promote  the 
union  of  SO2  with  O,  and  consequent  formation  of  SO3  and 
PbSO4),  sulphate  may  be  formed  almost  exclusively.  If  the 
temperature  is  then  raised,  the  tendency  of  PbSO4  to  react 
upon  the  undecomposed  PbS  rapidly  gets  stronger,  until  at  an 
orange  heat  this  takes  place  rapidly  and  fairly  completely, 
forming  the  one  single  gaseous  product,  SO2. 

PbS  +  PbSO4  =  2Pb  +  2SO2 

-  (20,200)  -  (215,700)  +2(69,260) 
Net  deficit,  97,380  Calories. 

PbS  +  2PbO  =  3Pb  +  SO2 

-  (20,200)  -  2(50,800)  +  (69,260) 
Net  deficit,  52,540  Calories. 

In  both  these  cases  the  well-known  phenomena  of  an  endo- 
thermic  reaction  are  manifest — the  high  temperature  and 
strong  firing  necessary,  and  the  formation  of  a  single  gaseous 
product  from  non-gaseous  materials,  assisting  the  reaction. 

The  reaction : 

PbS  +  3PbSO4  =  4PbO  +  4S02 
-  20,200  -  3(215,700)  +4(50,800)  +4(69,260) 
Net  deficit,  187,060  Calories, 


THE  METALLURGY  OF  LEAD.  583 

is  supposed  to  take  place  in  "  pot  roasting  "  where  excess  of  air 
is  blown  through  the  finely  divided  material,  but  it  is  too  endo- 
thermic  a  reaction  to  take  place  in  a  pot-roasting  operation  to 
more  than  a  very  subsidiary  extent. 

Oxidation  Refining. — Impure  lead  is  refined  or  "  softened  "  by 
oxidation  at  a  red  heat.  We  must  expect  a  great  deal  of  lead 
to  be  oxidized  in  this  operation,  simply  because  of  its  pre- 
ponderating mass,  but  the  impurities  present  will  oxidize  rela- 
tively faster  or  slower  than  lead  in  proportion  as  their  affinity 
for  oxygen  is  relatively  greater  or  less.  The  skimmings  or 
slags  obtained  during  softening  are  always  principally  com- 
posed of  lead  oxide,  PbO,  but  they  come  off  containing,  in 
order,  zinc,  tin,  antimony,  arsenic,  bismuth  and  small  quantities 
of  silver.  During  cupellation  down  to  silver,  which  is  con- 
tinued oxidation  until  all  the  lead  is  oxidized,  bismuth  oxide  is 
concentrated  in  the  last  parts  of  lead  oxide  formed,  which  may 
also  carry  silver  in  small  amount;  before  this  happens,  how- 
ever, the  lithage, formed  is  almost  chemically  pure. 

The  converse  of  these  oxidation  reactions  also  holds,  viz. : 
differential  reduction.  Taking  a  softening  skimming  rich  in 
antimony,  for  instance,  it  is  possible  by  mixing  it  with  a  small 
amount  of  reducing  agent,  such  as  carbon,  to  reduce  out  of  it 
all  the  silver  and  considerable  of  the  lead  which  it  contains 
without  reducing  much  antimony.  This  leaves  the  remaining 
unreduced  material  desilverized  and  poorer  in  lead,  or  richer 
in  antimony,  and  on  subsequent  reduction  of  this  by  excess  of 
reducing  agent  a  rich  antimony-lead  alloy  is  obtained.  The 
easy  reducibility  of  lead  oxide  is  complementary  to  the  slow 
oxidation  of  lead ;  both  facts  are  clear  from  the  heat  of  formation 
of  the  various  metallic  oxides,  and  both  are  extensively  utilized 
in  the  metallurgy  of  lead. 

THE  VOLATILITY  OF  LEAD. 

The  melting  point  of  lead  is  326°,  its  mean  specific  heat  in 
the  solid  state  0.02925  +  0.000019t,  heat  in  melted  lead  at  its 
melting  point  11.6  Calories,  latent  heat  of  fusion  4.0  Calories, 
heat  in  just  melted  lead  15.6  Calories,  specific  heat  in  the  liquid 
state  0.042 — and  approximately  constant — boiling  point  at 
normal  atmospheric  pressure  about  1,800°  C.,  latent  heat  of 
vaporization,  calculated  by  Trouton's  rule  (23  T)  47,680 


584  METALLURGICAL  CALCULATIONS. 

Calories  per  molecular  weight  =  230  Calories  per  kilogram, 
assuming  the  vapor  monatomic,  specific  gravity  of  vapor  103.5, 
referred  to  hydrogen  gas  at  the  same  temperature  and  pres- 
sure, or,  theoretically,  9.315  kg.  per  cubic  meter  at  0°  C.  and 
760  mm.  pressure,  as  a  standard  datum. 

The  question  of  the  volatility  of  lead  at  other  temperatures 
than  1,800°  C.  is  highly  important  in  the  smelting  of  lead  ores, 
yet  is  practically  an  unknown  quantity.  The  following  is  an  at- 
tempt to  calculate  these  data,  so  important  in  practical  metal- 
lurgy. 

The  vapor  tension  curve  of  mercury  is  known  for  very  low 
and  up  to  comparatively  high  pressures.  A  rule  has  been 
observed  between  mercury  and  water  vapor,  in  that  the  absolute 
temperatures  at  which  these  two  substances  have  the  same  vapor 
tensions  are  found  to  stand  in  the  ratio  1.7  to  1  through  a 
large  range  of  temperatures.  Since  lead  vapor  is  in  all  prob- 
ability monatomic,  like  mercury  vapor,  we  will  deduce  the 
vapor  tension  curve  of  lead  from  that  of  mercury,  using  the 
constant  ratio  derived  from  the  two  temperatures  at  which 
their  respective  vapors  have  atmospheric  tension,  viz.: 

TPb        1,800  +  273        2,073 
THg   "~  357  +  273  630 

The  following  table  gives  the  most  reliable  data  for  the 
vapor  tension  curve  of  mercury,  and  the  corresponding  data 
calculated  for  lead,  assuming  the  constant  ratio  3.3  between  the 
absolute  temperatures  at  which  they  have  the  same  vapor 
tension : 

Tension  of  Vapor  Mercury  Lead 

mm.ofHg.  C°  '                        C° 

0.0002  0  625 

0.004  33  735 

0.045  67  844 

0.28  100  954 

1.47  133  1,064 

5.73  167  1,173 

18.25  200  1,283 

50.  233  1,393 

106.  267  1,502 


THE  METALLURGY  OF  LEAD.  585 

Tension  of  Vapor  Mercury  Lead 

mm.  of  Hg.  C°  C° 

242.  300  1,612 

484.        Atmospheres.  333  1,722 

760.            =1.0  357  1,800 

849.            =1.1  367  1,841 

1588.            =2.1  400  1,951 

4.3  450  2,116 

8.0  500  2,280 

13.8  550  2,445 

22.3  600  2,609 

34.  650  2,774 

50.  700  2,938 

72.  750  3,103 

102.  800  3,267 

137.5  850  3,436 

162.'  880  3,525 

From  the  above  table  a  vapor  pressure  curve  of  mercury  and 
lead  can  be  constructed.  An  examination  of  the  data  shows 
that  lead  is  certainly  volatile  to  a  minute  extent  at  a  low  red 
heat,  and  that  a  current  of  inert  gas  passing  across  the  surface 
of  melted  lead  at  that  temperature  certainly  carries  away 
vapor  of  lead;  at  the  melting  point  of  silver  the  tension  is  only 
about  one-quarter  of  a  millimeter,  or  one  three  thousandth  of 
an  atmosphere,  yet  this  means  that  each  cubic  meter  of  inert  gas 
carries  off  one  three-thousandth  of  its  volume,  or  one-thirtieth 
of  1  per  cent  of  its  volume,  of  lead  vapor.  At  1,300°,  the  tem- 
perature of  a  commercial  zinc  retort,  or  of  a  lead  smelting 
furnace,  the  tension  is  about  one-fortieth  of  an  atmosphere, 
which  would  mean  that  any  other  gas  or  vapor  could  carry 
2.5  per  cent  of  its  volume  of  lead  vapor  with  it.  It  must  be 
remembered,  moreover,  that  such  gas  saturated  with  lead 
vapor  if  suddenly  cooled  does  not  deposit  its  excess  of  lead 
vapor  as  liquid  lead,  but  that  a  suspension  similar  to  hoar- 
frost almost  inevitably  results,  the  so-called  "  lead  fume," 
which  is  simply  molecularly  divided  liquid  or  solid  lead  carried 
in  suspension  by  a  current  of  gas.  The  lead  vapor  is  thus  al- 
most entirely  carried  out  of  the  furnace  without  condensation 
and  deposition. 

Looking  at  the  higher  pressures,  we  can  understand  why  an 


586  METALLURGICAL  CALCULATIONS. 

explosive  reaction  results  when  PbO  is  reduced  by  finely  di- 
vided aluminium,  the  "  alumino thermic  "  reaction.  The  im- 
mense heat  liberated  in  the  reaction, 

3PbO  +  2Al  =  3Pb  +  Al203, 

220,000  Calories,  raises  the  products  to  an  electric  furnace  tem- 
perature, to  some  3,000°  C.,  at  which  temperature  the  lead 
vapor  has  a  maximum  tension,  according  to  our  table,  of  60 
atmospheres.  No  wonder,  then,  that  when  Tissier  tried  this 
test  for  the  first  time,  in  1857,  using  a  piece  of  sheet  aluminium 
weighing  less  than  3  grams  (0.1  ounce),  "  le  creuset  a  e*te  brise 
en  mille  pieces  et  les  portes  du  fourneau  projetees  au  loin  " — 
the  crucible  was  broken  into  a  thousand  pieces  and  the  doors 
of  the  furnace  blown  to  a  distance. 

ROASTING  OF  LEAD  ORES. 

The  principal  operations  in  the  metallurgy  of  lead  are  the 
roasting  of  the  ore,  its  reduction  to  metal  and  the  refining  of 
the  crude  metal. 

The  chief  ore  of  lead  being  galena,  PbS,  the  operation  of 
roasting  it  in  air  converts  it  partly  into  PbO  and  partly  into 
PbSO4.  Since  PbS  is  easily  fusible,  and  is  volatile  per  se  at 
a  yellow  heat,  it  is  necessary  to  roast  carefully  and  slowly, 
avoiding  high  temperatures,  which  first  make  the  ore  sticky 
or  pasty  and  afterwards  fuse  it,  thus  practically  stopping  the 
roasting  reaction.  The  only  manner  in  which  rapid  roasting 
can  be  done  is  by  having  the  ore  mixed  with  so  much  infusible 
inert  matter,  like  lime,  that  the  globules  of  melted  sulphide 
cannot  run  together,  but  are  kept  isolated  and  continue  to 
oxidize  on  their  surfaces,  the  mass  being  meanwhile  kept  "  open  " 
or  porous  by  the  inert,  infusible  material,  to  allow  the  rapid 
passage  of  air  through  the  mass.  This  is  the  principle  of  "  pot 
roasting  " — the  most  radical  improvement  of  recent  years  in 
the  metallurgy  of  lead. 

Problem  123. 

A  Savelsberg  "  pot  roaster  "  treats  5  tons  of  ore  mixture, 
consisting  of  100  parts  lead  ore,  10  parts  quartzose  silver  ore, 
10  parts  spathic  iron  ore,  19  parts  limestone.  The  lead  ore  is 
galena  ore,  containing  78  per  cent  of  lead  and  15  per  cent  of 


THE  METALLURGY  OF  LEAD.  587 

sulphur;  the  silver  ore  may  be  called  silica  sand;  the  spathic 
iron  ore  FeCO3;  the  limestone  CaCO3.  The  mixture  is  mois- 
tened with  5  per  cent  of  its  weight  of  water.  Air  blast  is  kept 
constant  at  7  cubic  meters  per  minute,  blower  displacement  at 
15°  C.,  and  the  operation  lasts  18  hours,  leaving  2  per  cent  of 
sulphur  in  the  product,  which  may  be  assumed  as  undecom- 
posed  PbS.  Gases  produced  contain  approximately  10  per 
cent  of  SO2  and  5  per  cent  free  oxygen;  50  kg.  of  charcoal  is 
used  to  start  the  operation,  and  is  assumed  to  be  pure  carbon. 

Required  : 

(1)  The  weight.  of  product  and  its  percentage  composition. 

(2)  The  complete  analysis  of  the  gases  escaping. 

(3)  The  efficiency  of  delivery  of  the  blower. 

(4)  The  proportion  of  the  heat  generated  by  the  oxidation 
of  the  ore  which  is  absorbed  in  the  decomposition  of  the  lime- 
stone. 

(5)  The  proportion  of  the  heat  generated  in  the  pot  used 
in  evaporating  the  moisture  of  the  charge. 

(6)  The  proportion  of  the  heat  generated  in  the  pot  carried 
away  by  the  gases  at  an  average  temperature  of  300°  C. 

(7)  Make  a  heat  balance  sheet  of  the  whole  operation. 

Solution  : 

(1)  The  components  of  the  5-ton  (=5,000  kilos)  charge  are: 
Galena  ...........  3597  kg.  =  Pb       2806  kg.      S  :     540  kg. 

Silver  ore  ........   360    "    =  SiO2:     360    " 

Iron  ore  ..........   360    "    =  FeO  :     224    "    CO2:     136     " 

Limestone  ........   683    "    =  CaO:     382    "    CO2:    301     " 

Approximate  composition  of  product: 

ooq 

PbO  ...............  .......  .2806X         =  3023  kg. 


SiO2  .......................  =    360    " 

FeO  .......................  =    224    " 

CaO  .......................  =    382    " 

3989    " 

But,  since  product  contains  2  per  cent  of  sulphur  as  unde- 
composed  PbS,  the  above  weight  is  only  99  per  cent  of  the 


588  METALLURGICAL  CALCULATIONS. 

weight  of  the  roasted  ore,  because  the  lead  combined  with  this 
sulphur  has  been  calculated  above  to  PbO,  and  the  O  in  PbO 
is  only  half  the  weight  of  the  S  in  PbS.  The  above  weight  is 
short,  therefore,  by  an  amount  equal  to  one-half  the  weight  of 
the  sulphur  present,  and  since  the  latter  is  2  per  cent  of  the 
roasted  ore  the  above  weight  is  1  per  cent  short.  The  cor- 
rected weight  of  the  roasted  ore  is,  therefore, 

.      3,989 -J- 0.99  =  4,029kg., 

and  its  sulphur  content  81  kg.,  corresponding  to  521  kg.  of 
lead,  or  602  kg.  of  PbS.  This  leaves  present,  as  PbO,  2,806- 
521  =  2,285  kg.  of  lead,  equal  to  2,285X223/207  =  2,461  kg. 
of  PbO. 

Revised  composition  of  product: 

61  per  cent 


PbS      

602    "    =15 

SiO2  

360    "    =    9 

FeO  

224    "    =     5 

CaO.. 

.   382    "    =    9 

4029    "  (1) 

(2)  The  charge  gives  to  the  gases 

H2O  5000  X  0 . 05  =  250  kg. 

CO2  (50 X 3. 67) +  136  + 301  =  620    " 

S  540  -  81  =  459    " 

and  the  blast  gives  to  the  charge 

O  2461  -  2285  =  176  kg. 

The  volume  of  SO2  produced  from  459  kg.  of  sulphur,  which 
makes  918  kg.  of  SO2,  will  be,  at  standard  conditions, 


(0.09XTH  =  319  cubic  meters. 

The  free  oxygen  in  the  gases  is  half  the  volume  of  the  SO2, 
and,  therefore,  will  be  exactly  one-quarter  of  its  weight. 

The  volumes  of  H2O  and  CO2  present  in  the  gases,  at  standard 
conditions,  are 


THE  METALLURGY  OF  LEAD.  .  589 

H2O      250^-    /O.OQXy)     =  309  cubic  meters. 

CO2       620 -r-   (o.OQXy)     =  313      " 

The  nitrogen  present  is  that  in  the  air  used,  or  10/3  the 
weight  of  oxygen  in  the  air.  The  latter  is  the  weight  of  oxygen 
in  the  PbO  of  the  roasted  charge,  plus  the  oxygen  necessary  to 
burn  the  50  kg.  of  charcoal,  plus  the  oxygen  oxidizing  sulphur 
to  SO2  and  the  free  oxygen  in  the  gases. 

O  in  PbO =     176  kg. 

00 

O  to  burn  C 50Xff   =     133    " 

00 

O  to  burn  S. 459X^   =     459    " 

oZ 

Free  O2  in  the  gases =    230    " 


Total       998    ' 
N2  corresponding =  3327    " 

Composition  of  escaping  gases: 

Per  Cent. 

Nitrogen 3327  kg.  =  2609  cu.  meters  =  70.3 

Oxygen 230  "  =  159  "    "   =  4.3 

SO2 918  "  =  319  "    "   =8.6 

CO2 620  "  =  313  "    "   =  8.5 

H2O  vapor 250  "  =  309  "    "   =  8.4 


3709          100. 

(2) 

(3)  Oxygen  delivered  by  blast 998  kg. 

Nitrogen  in  the  blast 3327    " 

Air  delivered 4325    " 

Volume  at  0°  =  4325-1-1.293  =  3345  cu.  meters. 

ooo 

Volume  at  15°  =  3345  X  ~  =  3529    " 

Z  i  o 


590  METALLURGICAL  CALCULATIONS. 

Displacement  of  blower  at  15° 

7X60X18  =  7560    " 

oc  on 

Efficiency  of  blower       =—  =  0.467  =  46.7  per  cent.      (3) 
«oou 

(4)  The  heat  generated  in  the  oxidation  of  the  ore  consists 
of  the  heat  of  formation  of  the  PbO  formed,  plus  that  of  the 
SO2,  and  less  that  of  the  PbS  decomposed.  This  is  not  the 
complete  heat  balance  of  the  whole  operation,  but  simply 
that  concerned  with  the  oxidation  of  the  galena. 

Heat  of  oxidation  of  Pb  to  PbO 

2285  kg.  Pb.X245        =  +    559,825  Cal. 

Heat  of  oxidation  of  S  to  SO2 

459kg.  SX  2164          =+     993,275     " 

Heat  of  formation  of  PbS  decomposed 

2285kg.  PbX  98  =-    223,930     " 


Roasting  heat  =  +1,329,170     " 

The  decomposition  of  the  limestone,  producing  301  kg.  of 
CO2,  absorbs 

301X1,026  =  308,825  Cal. 

The  proportion  of  the  roasting  heat  thus  absorbed  is 

0.232  =  23.2  per  cent.  (4) 


(5)  The  5-ton  charge  is  moistened  with  5  per  cent  of  its 
weight  of  water,  or  250  kgs.,  in  order  to  keep  down  the  tem- 
perature. This  absorbs,  simply  in  becoming  vapor, 

250X606.5  =  151,625  Cal., 

Which,  on  the  heat  generated  in  the  roasting  operation,  is 
151,625 


1,329,170 


-  0.114  -  11.4  per  cent.  (5) 


(6)  If  the  gases  pass  away  at  an  average  temperature  of  300°, 
and  the  air  comes  in  at  an  average  temperature  of  15°,  the 


THE  METALLURGY  OF  LEAD.  591 

gases  cool  off  the  pot  to  the  extent  of  the  difference  between 
these  two  heat  capacities: 

Heat  in  the  air  entering: 

3345  m3X [0.303  + 0.000027(15)]  X 15  =  15,225  Cal. 

Heat  in  gases  escaping: 

™    ^n^i-  X [0.303  +  0.000027(300)]      -    861.1  Cal.  per  1°. 
u       loy  m  ) 

SO2     319m3    X[0.36  +  0.0003(300)]  =     143.6 

CO2     313m3    X [0.37 +  0.00022(300)]          =     136.5 
H2O    309m3    X [0.34 +  0.00015(300)]          =     119.0 

1260.2 

Total  =  1260.2X300  =  378,060  Cal. 

Proportion  of  the  total  roasting  heat  thus  carried  out: 
362,835 


1,329,070 


0.280  =  28.0  per  cent.  (6) 


(7)                       Heat  Available.  Calories. 

Sensible  heat  of  air  blast,  at  15° 15,225 

Combustion  of  charcoal 405,000 

Net  heat  generated  by  the  roasting 1,329,170 


Total 1,749,395 

Heat  Distribution. 

Sensible  heat  in  the  hot  gases 378,060  =  22  per  cent. 

Decomposition  of  carbonates : 

Calcium  carbonate 308,825  =  18  " 

Iron  carbonate. . 76,980  =4  " 

Evaporation  of  moisture 151,625  =9  " 

Sensible  heat  of  product,  at  400°. . . .    400,000  =  23  " 

Loss  by  radiation  and  conduction. . .     433,905  =  25  " 


1,749,395 

REDUCTION  OF  ROASTED  ORE. 

If  the  roasted  ore  is  reduced  in  reverberatory  furnaces,  the 
chances  are  that  the  undecomposed  sulphide  it  contains  will 
react  with  the  oxide  or  sulphate,  forming  metal  and  SO2,  and 


592  METALLURGICAL  CALCULATIONS. 

thus  eliminating  almost  all  of  the  sulphur  left  in  the  roasted 
ore;  this  results  in  the  formation  of  no  matte,  or  at  most  of 
very  little  rich  matte.  If  the  roasted  ore  is  reduced  in  shaft 
furnaces,  the  carbonaceous  fuel  and  strong  reducing  atmos- 
phere of  the  same,  tend  to  reduce  oxides  to  metal  and  sul- 
phates to  sulphides  before  they  reach  the  temperature  at  which 
they  can  react  on  sulphide,  thus  cutting  out  very  largely  the 
double  reaction,  and  throwing  into  matte  the  larger  part  of 
the  sulphur  left  in  the  roasted  ore.  This  is  highly  advan- 
tageous if  the  ore  has  copper  in  it,  even  if  only  in  small  amount, 
as  the  matte  will  concentrate  the  copper  in  it,  is  easily  sep- 
arated from  the  metal  and  the  slag,  and  forms  a  valuable  by- 
product. Since  most  lead  ores  contain  some  copper,  enough 
to  make  it  pay  to  save  it,  the  roasting  of  these  is  never  pushed 
to  completion,  and  the  reduction  in  low  shaft  furnaces,  run 
at  moderate  temperature,  is  pursued  with  the  object  of  pro- 
ducing a  copper-iron  matte.  Any  arsenic  or  antimony  left  in 
the  roasted  ore  is  likely  also  to  form  spiess,  a  compound  of 
arsenic  and  antimony  with  iron,  copper,  nickel,  cobalt,  lead 
and  silver,  which  is  heavier  than  matte,  but  lighter  than  lead, 
and  separates  out  in  the  cooling  pots  between  these.  It  is  a 
highly  undesirable  product,  because  of  its  complexity  and  the 
difficulty  of  satisfactorily  and  cheaply  separating  its  constitu- 
ents; it  is  always  to  be  advised  to  remove  arsenic  and  anti- 
mony as  completely  as  possible  in  the  roasting  operation,  even 
at  the  expense  of  removing  too  much  sulphur,  and  then  to  supply 
the  sulphur  needed  in  the  smelting  operation  by  mixing  with 
the  dead-roasted  ore  some  raw  sulphide  ore  free  from  arsenic 
and  antimony,  if  such  can  be  obtained. 

In  calculating  the  charge  for  such  a  smelting  operation,  the 
roasted  ore  must  be  charged  with  such  amounts  of  iron  ore  and 
limestone  as  will,  together  with  the  gangue  of  the  ore  and  the 
ash  of  the  fuel  used,  make  an  easily  fusible  slag.  To  produce 
such  a  slag  is  of  fundamental  importance,  since  it  must  melt 
and  become  thinly  liquid  at  the  moderate  temperature  neces- 
sarily prevailing  in  a  lead  furnace.  Experience  has  shown 
that  a  typical  lead  slag  will  contain  about  30  per  cent  SiO2,  40 
per  cent  FeO  and  20  per  cent  CaO,  with  10  per  cent  allowed 
for  other  ingredients,  such  as  APO3,  etc.,  or  in  more  general 
terms,  that  SiO2,  FeO  and  CaO  are  best  present  in  the  propor- 


THE  METALLURGY  OF  LEAD.  593 

tions  30  :  40  :  20.  The  other  limitations  are  that  a  certain 
amount  of  hand-picked  slag  is  always  returned  to  the  furnace 
for  re-smelting  and  purification  from  matte,  and  that  the  fuel, 
coke,  cannot  melt  more  than  a  certain  amount,  say  seven 
times  its  weight  of  inert  material.  With  these  conditions  in 
mind,  we  will  work  out  a  typical  smelting  problem,  the  data 
for  which  are  taken  from  Hofman's  Metallurgy  of  Lead. 

Problem  124. 

A  mixture  of  lead  carbonate  ore  and  galena  (which  repre- 
sents pretty  closely  a  partially  roasted  sulphide  ore)  is  smelted 
with  the  addition  of  iron  ore  and  limestone.  The  charges  for 
the  furnace  consist  of  1,000  pounds  of  burden  (material  to  be 
smelted)  and  150  pounds  of  coke  (containing  10  per  cent  ash), 
and  the  1,000  pounds  consist  of  100  pounds  return  slags  and 
900  pounds  of  lead  ore,  iron  ore  and  limestone.  The  percentage 
composition  of  these  materials  is :  • 

Lead  Ore.  Iron  Ore.     Limestone.  •  Ash  of  Coke. 

SiO2 32.6  4.3  2.7  40.3 

FeO 14.8  72.4  4.5  26.5 

MnO 4.3  1.7 

CaO 2.2  3.1  37.3  6.9 

MgO 5.3  11.9  2.4 

APO3 :..       2.5  ....  ....  20.4 

BaO 1.5 

ZnO 2.4  The  iron  present  in  these  mate- 

S 4.4         rials   is    really   present   mostly   as 

As 0.5         Fe2O3  and  only  partly  as  FeO,  but 

Pb 20 . 7         the  analysis  is  expressed  usually  as 

Cu 2.9         FeO  in  order  to  facilitate  the  slag 

Ag 0. 17       calculations.  The  analyses,  as  given, 

are  not  expected  to  add  up  to  100. 

In  making  the  calculations,  assume  that  the  slag  to  be  pro- 
duced contains  SiO2:  FeO:  CaO  in  the  ratio  30  :  40  :20;  that 
the  FeO  here  expressed  includes  the  MnO,  calculated  to  its 
FeO  equivalent,  and  the  CaO  includes  the  MgO,  BaO,  and 
ZnO  calculated  to  their  CaO  equivalent.  Assume  also  that 
all  the  ZnO  of  the  charge  goes  into  the  slag,  all  the  sulphur 
into  matte,  all  the  arsenic  into  speiss,  of  the  formula  Fe5As, 
all  the  silver  and  lead  into  the  lead  bullion. 


594  METALLURGICAL  CALCULATIONS. 

Requirements  : 

(1)  The  proportions  of  lead  ore,  iron  ore  and  limestone  to 
be  used  in  the  900  pounds  of  these  to  a  charge. 

(2)  The  weight  and  composition  of  the  slag  formed. 

(3)  The  weight  and  composition  of  matte  formed. 

(4)  The  weight  and  composition  of  speiss  formed. 

(5)  The  weight  and  composition  of  lead  bullion  formed. 

(6)  A  balance  sheet  of  the  essential  materials  entering  and 
leaving  the  furnace. 

Solution  : 

(1)  There  are  really  only  two  unknown  weights  to  be  de- 
termined, for  the  sum  of  the  three  weights  required  is  to  be 
900.  Similarly,  the  ratio  30  :  40  :  20  really  gives  two  condi- 
tions to  be  fulfilled,  since  there  are  practically  two  ratios  to 
be  worked  to.  The  simplest  method  of  solution  is,  undoubt- 
edly," the  algebraic  one,  letting  x  and  y  represent  two  of  the 
ores,  900  —  (x  +  y)  the  other,  and  then  working  out  the  weights 
of  SiO2,  FeO  and  CaO  in  the  slag,  expressed  in  terms  of  x  and  y. 
The  ratio  30  :  40  :  20  then  gives  us  two  independent  equations, 
containing  the  two  unknowns,  and  the  problem  is  solved. 

Let  x  =  weight  of  lead  ore. 

y  =  weight  of  iron  ore.  . 

900  —  x  —  y  =  weight  of  limestone. 

15  =  weight  of  ash  of  coke. 
100  =  weight  of  return  slags. 

Calculation  of  FeO  going  into  Slag. 

Sulphur  in  100  parts  ore  4.4  Ibs. 

Copper  in  100  parts  ore  2.9     " 

Sulphur  united  with  Cu  to  form  Cu2S  0.7     " 

Sulphur  left  over  to  form  FeS  3.7     " 

Iron  needed  to  form  FeS  =  3.7  x||  =     6.5     " 

o5a 

72 

FeO  corresponding  to  this  Fe  =  6.5X^  =     8.4     " 

oo 

Arsenic  in  100  parts  of  ore  0.5     " 


Iron  required  to  form  Fe5As  =  0.5X-^~  =      1.9     " 

i  O 


THE  METALLURGY  OF  LEAD.  595 

72 

FeO  corresponding  to  this  Fe  =  1.9X-™  =     2.4  Ibs. 

oo 

Total  FeO  disappearing  in  matte  and  speiss         =10.8     " 
FeO  left  over  to  go  into  slag  =  14.8-  10.8  =     4.0     " 

Therefore  x  parts  of  lead  ore  contributes  to  the  slag: 

SiO2 0.326  x 

FeO 0.040  x 

MnO 0.043  x  =  0.044  x  FeO. 

CaO 0.022  x 

MgO 0.053  x  =  0.074  x  CaO. 

APO3 0.025  x 

BaO 0.015  x  =  0.006  x  CaO. 

ZnO 0.024  x  =  0.017  x  CaO. 

'  The  FeO  equivalent  of  MnO  is  72/71  the  MnO;  the  CaO 
equivalent  of  MgO  is  56/40  the  MgO,  of  BaO  is  56/141  the 
BaO,  of  ZnO  is,  56/81  the  ZnO.  These  ratios  are  the  chem-' 
ically  equivalent  values  of  these  oxides,  as  taken  from  their 
molecular  weights.  Adding  all  these  together,  our  lead  ore 
contributes  to  the  slag  the  equivalent  of 

SiO2  0.326  x,  FeO  0.084  x,  CaO  0.119  x. 
The  y  parts  of  iron  ore  similarly  contributes  to  the  slag: 

SiO2  0.043  y,  FeO  0.741  y,  CaO  0.031  y. 
The  900  —  x  —  y  parts  of  limestone  contributes  likewise : 

SiO2  0.027 (900- x-y) 

FeO  0.045  (900 -x-y) 

CaO  0.373(900 -x-y) 

MgO  0.1 19  (900- x-y)  =  0.167(900  -  x  -  y)  of  CaO. 

Therefore,  total  contribution  of  the  limestone  to  the  slag: 

SiO2  0.027(900  -  x  -  y),  FeO  0.045(900  -x-y), 
CaO  0.540  (900- x-y). 

Contribution  of  ash  of  fuel  to  the  slag,  in  similar  manner: 
SiO2  6.1  FeO  4.0  CaO  1.4 


596  METALLURGICAL  CALCULATIONS. 

Adding  all  these  together,  we  have  in  the  slag: 

SiO2  .........................   30.4  +  0.299  x  +  0.016  y 

FeO  .........................   44.5  +  0.039  x  +  0.696  y 

CaO  .........................  487.6  -0.421  x  -0.509  y 

According  to  assumption,  these  ingredients  as  thus  summated 
should  bear  the  relations  30  :  40  :  20.     We  therefore  have 

OQ 

30.4  +  0.299  x  +  0.016  y  =  —  (487.6  -  0.421  x-  0.509  y) 


44.5  +  0.039  x  +  0.696  y  =  ^  (487.6  -  0.421  x-  0.509  y) 

Z(j 

Whence  x  =  lead  ore        =  523  Ibs. 

y  =  iron  ore        =  274    " 

900  -  x  -  y  =  limestone     =  103    "  (1) 

(2)  With  the  weights  of  materials  used,  as  calculated,  we  can 
calculate  the  weights  of  the  ingredients  of  the  slag  as: 

Lead  Ore.  Iron  Ore.  Limestone.  Coke  Ash.  Total. 

SiO2  ..........      170.5       11.8             2.8             6.1  191.2 

FeO  ..........       20.9     198.4             4.6             4.0  227.9 

MnO  .........       22.5         4.7           ........  27.2 

CaO  ..........        11,5         8.5           38.4             1.0  59.4 

MgO  ..........       27.7       ____            12.3             0.4  40.4 

A12O3  .........        13.1       ........             3.1  16.2 

BaO  ..........         7.8       ............  7.8 

ZnO..                       12.6  12.6 


582.7 
Percentage  Composition  and  Check  on  Ratio. 

SiO2 32.8percent 

FeO. 39.0 

MnO 4.7        "         =  4.8  per  cent  FeO. 

CaO 10.2 

MgO 6.9        "         =  9.7        "       CaO. 

APO3 2.8 

BaO 1.3        u         =  0.5        "       CaO. 

ZnO 2.2        "         =  1.5        «       CaO. 

Summated  SiO2  :  FeO  :  CaO 

=  32.8  :  43.8  :  21.9  =  30  :  40  : 20 


THE  METALLURGY  OF  LEAD.  597 

(3)  Sulphur  in  523  Ibs.  of  lead  ore  =  23.0  Ibs. 
Copper  in  523  Ibs.  of  lead  ore       :  =  15.2    " 
Sulphur  to  form  Cu2S  with  this  Cu  =     3.8    " 
Sulphur  to  form  FeS  =  23;0  -  3.8  =  19.2    " 

Fe  to  form  FeS  =  19.2  X  ~|  =  33.6    " 

FeS  in  matte    =19.2  +  33.6  =52.8    " 

Cu2S  in  matte  =  15.2  +  3.8  =  19.0    " 

Total  weight  of  matte  =  71.8    " 

(3) 
Composition  of  matte 

Cu2S    =  26.5  per  cent  =  Cu     =21  per  cent. 
FeS     =  73.5        "  Fe     =  47 

S       =22        "  (3) 

(4)  Arsenic  in  523  Ibs.  of  lead  ore  =    2.6    " 

9&0 

Fe  to  form  Fe5As  =2.6X  -~  =    9.7    « 

7o 

Weight  of  speiss  formed  =  12.3    " 

(4) 
Composition :  Fe  79  per  cent 

As  21        "  (4) 

(5)  Lead  in  523  Ibs.  of  lead  ore  =  108.3  Ibs. 
Silver  in  523  Ibs.  of  lead  ore  =      0.9    " 


109.2    " 

(5) 
Composition:  Pb  99.2  per  cent. 

Ag    0.8        "  (5) 

(6)   Balance  sheet  of  materials  entering  and  leaving  per  1,000 
Ibs.  of  burden: 

Charges.              Bullion.  Matte.     Speiss.     Slag.  Gases. 
Lead  Ore   523 

SiO2          170.5     170.5         

FeO            77.4     ...    Fe  33.6  Fe  9.7  Fe  O20.9  O  13.2 

MnO  22.5     ...  22.5         

CaO  11.5  11.5 


598 


METALLURGICAL  CALCULATIONS. 


Charges. 
MgO 
APO3 

Bullion. 
27.7     ... 
13.1      ... 

Matte 

BaO 

7.8     ... 

.... 

ZnO 

12.6      ... 

.... 

S 

23.0      ... 

23.0 

As 

2.6      ... 

.... 

Pb 

108.3  108.3 

.... 

Cu 

15.2      ... 

15.2 

Ag 

0.9     0.9 

Iron  Ore 

274 

SiO2 

11.8     ... 



FeO 

198.4     ...    , 

. 

MnO 

4.7     ... 

.... 

CaO 

8.5     ... 

.... 

Limestone 

103 

SiO2 

2.8     ... 

.  .  .  • 

FeO 

4.6     ... 

.... 

CaO 
MgO 

38.4     ... 
12.3     



o 

Ash  of  Fuel 

15 

SiO2 

6.1      ... 

^ 

FeO 

4.0     ... 

.... 

CaO 

1.0     ... 

MgO 
A12O3 

0.4     ... 
3.1     ... 

.... 

Return  Slag 

100     ... 



Speiss.     Slag.         Gases. 

27.7         

13.1 

7.8 

12.6 


2.6 


11.8 

198.4 

4.7 

8.5 

2.8 

4.6 

38.4 

12.3 

6.1 
4.0 
1.0 
0.4 
3.1 
100.0 


109.2       71.8         12.3       682.7 


13.2 


THE  ELECTROMETALLURGY  OF  LEAD. 

As  far  as  the  present,  the  use  of  electrical  methods  in  the 
metallurgy  of  lead  has  been  confined  to  the  Salom  process  of 
cathodic  reduction  of  galena  and  the  Betts  process  of  refining 
crude  lead  bullion.  Electrothermic  methods  of  smelting,  and 
even  of  roasting,  are  among  future  possibilities,  as  are  also  the 
leaching  by  suitable  solvents  and  electrical  precipitation  of 
lead  from  the  solutions,  but  they  have  not  been  commercially 
practiced. 

The  Salom  process  puts  the  powdered  galena  on  a  "  hard- 


THE  METALLURGY  OF  LEAD.  599 

lead  "  plate,  in  a  solution  of  dilute  sulphuric  acid,  and  using 
a  "  hard-lead  "  anode,  passes  a  strong  current  through,  reducing 
PbS  in  situ  to  Pb  with  formation  of  H2S  (with  some  hydro- 
gen) at  the  cathode  and  oxygen  at  the  anode.  It  was  run  . 
commercially  on  a  fair  scale  at  Niagara  Falls.  For  further 
details  reference  may  be  made  to  Transactions  American  Elec- 
trochemical Society,  Vol.  I,  p.  87;  II,  p.  65;  IV,  p.  101. 

The  Betts  process  consists  in  using  impure  lead  as  anode 
in  a  solution  of  lead  fluo-silicate,  PbF2.  SiF4,  strongly  acid 
with  HF.  The  refining  plant  and  its  operation  are  quite  sim- 
ilar to  a  copper  refining  plant.  The  cathodes  are  sheet  steel, 
greased,  and  the  dense  sheet  lead  deposited  is' stripped  off  from 
time  to  time. 

Problem  125. 

In  a  Salom  apparatus  powdered  galena,  density  of  powder 
3.5,  is  placed  in  a  layer  0.5  mm.  thick  on  a  revolving  lead  table 
having  an  effective  treatment  area  of  2.25  square  meters. 
Current  density  330  amps,  per  square  meter  of  cathode  surface. 
Time  of  treatment  of  the  layer  90  minutes.  Electrolyte  dilute 
H2SO4.  Resistance  of  cell  0.001  ohm.  Heats  of  formation: 
(Pb,  S)  20,300;  (H2,  S)  4,800  (as  gas);  (H2,  O)  69,000.  Assume 
reduction  to  Pb  complete. 

Required : 

(1)  The  efficiency  of  application  of  the  amperes  passing  to 
the  reduction   of   the  galena,   and   the   percentage   of  ampere 
efficiency  lost  by  the  evolution  of  hydrogen. 

(2)  The  average  composition  of  the  gases  coming  from  the 
cell  and  their  volume  per  hour,  at  normal  pressure  and  20°  C., 
assuming  the    electrolyte  saturated  with   H2S,    H2  and  O2    at 
starting. 

(3)  The  working  voltage   absorbed,   if  the   efficiency  of  re- 
duction of  galena  were  100  per  cent. 

(4)  The  working  voltage  if  the  cell  were  kept  running  after 
all  the  galena  was  reduced. 

(5)  The  average  working  voltage  of  the  cell  in  actual  opera- 
tion and  the  distribution  of  this. 

(6)  The  proportion  of  the  power  required  by  the  cell  which 
could  be  generated  by  burning  the  gases  produced  in  a  gas  engine 
(if  such  were  possible)  at  a  thermo-mechanical  efficiency  of  100 
per  cent. 


600  METALLURGICAL  CALCULATIONS, 

(1)   Current  used  2.25X330  =     742  amperes. 

Lead  theoretically  reducible  in  90  minutes 

0.00001035X103.5X60X90X742.5       =  4295  grams. 

Galena  in  the  apparatus  : 

2.25  X  10,000  X  0.05  X  3.5  =  3938      " 

Lead  under  treatment: 

907 
3938X^  =  3410      « 

Ampere  efficiency  of  the  treatment: 

=  0.794  =  79.4  per  cent. 


Per  cent  of  amperes  evolving  hydrogen  =  20.6  (1) 

(2)   If  all  the  current  evolved  IPS,  the  gases  evolved  would  be: 
2H2S         O2 
2  parts     1  part. 

If  all  the  current  evolved  H2,  the  gases  would  be: 
2H2  O2 

•  2  parts     1  part. 

If  79.4  per  cent  of  the  current  evolves  H2S  and  20.6  per  cent 
hydrogen,  the  average  gases  evolved  will  be: 
H2S       1.588  parts  =  62.93  per  cent. 
H2         0.412      "      =  13.73 
O2         1.000      "      =  33.33        "  (2) 

The  volume  evolved  per  hour  is  found  as  follows: 

Oxygen  produced  by  1  ampere  hour 

0.00001035X8X60X60     =       0.29808  grams. 

Produced  by  742.5  amperes,  in  one  hour 

0.29808X742.5  =  221.3 

Volume  at  760  mm.  and  0°  C. 

221.3-1.44  =  153.7          litres. 

Volume  of  gases  produced  per  hour 

153.7X3  =  461. 

=       0.461      cubic  meter 
Volume  at  20° 

0.461  X27:Lt2°=      0.495          «  (2) 

Zi  o 


THE  METALLURGY  OF  LEAD.  601 

(3)  Voltage  for  ohmic  resistance 

0.001X742  =  0.742  volts. 
Chemical  work  done  if  only  H2S  is  formed,  per  formula 

PbS  +  H2O  =  H2S  +  O  +  Pb 

-  20,300  -  69,000  +4,800  =  -  84,500  Cai. 

Per  chemical  equivalent  concerned  =  —  42,250     " 

Voltage  absorbed  in  chemical  work 

42,250-^23,040  =  1.83  volts 
Total  voltage  drop  in  the  cell  1.83  +  0.74      =2.59      "     (3) 

(4)  Chemical  work  done  if  only  H2  is  liberated  =  69,000  Cal. 
Voltage  absorbed  in  chemical  work 

^2^23,040=      1.49  volts. 
Total  voltage  drop  in  the  cell  =      2.23  (4) 

(5)  Voltage  absorbed  in  chemical  work  if  79.4  per  cent  of  the 
current  evolves  H2S  and  20.6  per  cent  H2: 

(1.83X0.794) +  (1.49X0.206)   =  1.76  volts. 
Total  voltage  drop,  in  average  running          =  2.50      "  (5) 

Logically,  the  1.76  volts  absorbed  in  chemical  decomposition  in 
average  running  is  properly  calculated  thus,  from  the  energy 
evolved: 

(42,250X0.794)  +  (34,500x0.206) 
23,040 

(6)  Heat  of  combustion  of  1  cubic  meter  of  the  average  gas : 

H2S      0.5293  m3X5,513    =  2918  Cal. 
H2         0. 1373  m3X  2,614   =     359     " 


Sum  =  3277 

Calorific  power  of  gas  produced  per  hour: 
3277X0.461  =  1511  Calories. 

Watt  hours  producible  from  this  at  100  per  cent  thermo-mechan- 
ical  efficiency : 

1511^0.860  =  1757  watts. 


602  METALLURGICAL  CALCULATIONS. 

Current  used  by  the  cell,  average  running: 

742.5X2.50  =  1856  watts. 
Proportion  theoretically  regainable: 

=  0.95  =  95  per  cent.  (6) 


The  reason  why  more  power  is  theoretically  regainable  than 
is  used  in  doing  chemical  work  in  the  cell,  is  that  the  gas  engine 
burns  the  sulphur  of  the  PbS  ultimately  to  SO2,  and  this  is 
the  source  of  great  energy,  while  only  a  relatively  small  amount 
of  energy  was  necessary  to  convert  the  PbS  into  H2S  ready  for 
combustion. 

Problem  126. 

Lead  bullion  refined  by  the  Betts  process  is  96.73  per  cent 
lead,  and  the  refined  lead  is  practically  pure.  The  anodes  are 
1.5  inches  thick,  and  weigh,  with  lugs  for  suspension,  275  pounds. 
They  are  left  in  the  tanks  an  average  of  nine  days,  running 
with  135  amps,  per  plate,  whose  active  surface  is  1,700  square 
inches.  Space  between  anode  and  cathode  1J  inches,  specific 
resistance  of  solution  10  ohms.  Each  tank  has  twenty-two 
anodes  and  twenty-three  cathodes,  and  produces  an  average 
of  545  pounds  of  lead  per  day.  Power  costs  $50  per  electric 
horsepower-year,  as  delivered  to  the  tanks.  Drop  per  tank, 
due  to  resistance  of  contacts,  0.15  volt;  specific  gravity  of  pure 
lead,  11.35. 

Required  : 

(1)  The  weight  of  lead  theoretically  deposited  by  the  cur- 
rent per  day. 

(2)  The  ampere  efficiency  of  the  deposition. 

(3)  The  voltage  drop  per  tank. 

(4)  The  amount  of  anode  scrap  to  be  remelted  in  percent 
of  the  total  anode  weight. 

(5)  The  average  thickness  of  the  cathode  deposit  per  day. 

(6)  The  power  costs  per  ton  of  impure  lead  refined. 

Solution: 

(1)  Lead  theoretically  deposited  by  1  ampere  per  day 

0  .  00001035  X  103  .  5  X  60  X  60  X  24  =     92.55  grams. 


THE  METALLURGY  OF  LEAD.  603 

Per  anode  per  day,  dissolved 

92.55X135^1000  =     12.5  kg. 
=    27.5  Ibs. 
Per  tank,  per  day,  deposited 

27.5X22=605     "  (1) 

(2)   Efficiency  of  deposition 


^J  =  0.90  =  90  per  cent.  (2) 

DUO 

(3)  Amperes  used  per  plate  =  135 

Current  density  per  square  inch 

=  135^1700=      0.0794  amperes 

Current  density  per  square  centimeter 

=  0.0794-6.25  =      0.0127 

Distance  of  plates  apart  =       1.17  inches 

=      2.93  cm. 
Resistance  of  1  cm.  cube  of  electrolyte     =•-     10       ohms. 

Voltage  drop  in  electrolyte 

10  X  2 . 93  X  0 . 0127  =      0.37  volt. 

Voltage  drop  in  connections  =       0.15 

Total  voltage  drop  per  tank  =      0.52      "  (3) 

(4)  Dissolved  from  anode  in  9  days 

545 -v- 22X9  =  223       Ibs. 

Anode  corroded  in  9  days,  assuming  slime  to  fall  off  it 

223 -=-0.9673  =  231       Ibs. 
Anode  scrap  =  275  -  231  =     44 

44 

Percentage  of  anode  scrap  7^  =  0.16  =     16    per  cent. 


(5)  Deposit  of  lead  one  side  of  a  cathode,  per  day 

545^-44  =     12.4    Ibs. 
=      5.636  kg. 
Area  of  1  side  of  a  cathode  =  850      sq.  inches. 

=  5312     sq.  cm. 
Deposit,  per  day,  per  sq.  cm. 

5636^5312  =      1.06    grams. 


604  METALLURGICAL  CALCULATIONS. 

Volume  of  this  deposit  1 . 06  -v- 1 1 . 35  =       0.093  cu.  cm. 

Therefore  thickness  deposited  per  day       =      0.93    mm.         (5) 

(6)   Bullion  refined,  per  tank,  per  day 

545 -=-0.9673  =    563       Ibs. 

Power  required  per  tank          2970X0.52  =  1544       watts. 

2.06  e.h.p. 
Cost  of  power  per  year  =  50X2.06  =     103.      dollars. 

per  day  =  103^-365  =        0.28 

9nnn 
per  ton  of  lead  treated  0.28X^  =        0.99        *        (6) 


CHAPTER  III. 
THE  METALLURGY  OF  SILVER  AND  GOLD. 

The  processes  most  used  for  the  extraction  of  silver  and 
gold  from  their  ores  are  the  smelting  of  the  ores  with  copper 
ores  to  matte  and  eventually  blister  copper,  followed  by  electro- 
lytic refining  of  the  latter,  or  smelting  with  lead  ores,  followed 
by  desilverization  of  the  resulting  lead  bullion,  cupellation  of 
the  rich  lead,  or  electrolytic  refining  of  the  same.  In  all  of  these 
cases  the  silver  or  gold  is  a  comparatively  minute  constitutent 
of  the  ore,  intermediate  products  and  final  metal,  so  that  the 
metallurgy  of  silver  or  gold,  after  these  methods,  up  to  the 
production  of  impure  silver  or  gold,  or  silver  or  gold  bullion,  is 
practically  the  metallurgy  of  copper  or  lead — both  of  which 
subjects  we  have  considered  at  length. 

When  we  obtain  products  in  which  the  silver  or  gold  is  the 
chief  constituent,  we  approach  a  condition  in  which  calculations 
upon  the  precious  metal  contained  may  be  based,  but  here 
again  there  are  not  the  same  conditions  of  economy  so  promi- 
nent in  the  metallurgy  of  lead  or  copper.  It  does  not  matter 
greatly,  for  instance,  whether  we  use  50  cents'  worth  of  one 
fuel  or  a  dollar's  worth  of  another  in  melting  1,000  ounces 
of  silver  worth  over  $500;  the  question  of  which  fuel  is  the 
cleanest  or  most  convenient  is  of  greater  importance  than  the 
cost  of  the  fuel. 

The  object  of  this  introduction  is  to  show  that,  until  it  comes 
to  handling  nearly  pure  silver  or  gold,  close  calculations  as  to 
amounts  of  reagents,  running  of  furnaces,  etc.,  are  not  prac- 
ticable for  silver  or  gold,  per  se. 

ELECTROLYTIC  REFINING  OF  SILVER  BULLION. 

The  parting  of  gold  and  silver  by  acids  is  rapidly  giving 
place  to  the  cleaner  and  less  expensive  electrolytic  separation. 
The  bullion  may  be  of  quite  variable  composition.  The  follow- 
ing analyses  of  two  samples  give  an  idea  of  this  variation. 

605 


606  METALLURGICAL  CALCULATIONS. 

Cupelled  Silver.  Impure  Silver. 

Ag 98.69  74.08 

Pb 1.09  3.71 

Cu 0.12  20.23 

Fe 0.09  1.01 

Au 0.0023  0.05 

The  electrolytic  refining  proceeds  easiest  with  the  nearly 
pure  silver,  because  the  impurities  going  into  solution  accumu- 
late much  more  slowly,  and  the  solution  therefore  requires 
purifying  less  frequently.  On  the  other  hand,  the  solution  of 
a  large  amount  of  copper,  iron  or  lead  adds  electromotive  force 
to  the  circuit,  since  silver  only  is  deposited,  and  thus  decreases 
the  electrical  work  to  be  done. 

Problem  127. 

Two  varieties  of  silver  bullion  are  refined  electrolytically 
carrying 

No.  1.  No.  2. 

Ag 98.69  74.08 

Pb 1.09  3.71 

Cu 0.12  20.23 

Fe 0.09  1.01 

Au 0.01  0.05 

The  slimes  from  each  carry  in  percentages: 

No.  1.  No.  2. 

Ag 60  55 

Pb 10  5 

Cu 5  15 

Fe '....         3  5 

Au 22  20 

In  each  case  all  the  gold  goes  into  the  slimes.  A  current 
density  of  200  amperes  per  square  meter  of  electrode  surface  is 
used,  and  the  plates  are  2  centimeters  thick.  Specific  gravity 
of  No.  1,  10.15;  of  No.  2,  9.79.  There  is  anode  scrap  equal  to 
12  per  cent  of  the  weight  of  the  plates.  The  working  space 
between  the  electrodes  is  4  cm.  'total  current  220  amperes  per 
tank,  specific  resistance  of  electrolyte  20  ohms.  Deposited 
silver,  19.85  kg.  per  tank  per  day. 


THE  METALLURGY  OF  SILVER  AND  GOLD. 


607 


Required  : 

(1)  The  ampere  efficiency  of  the  deposition  of  silver. 

(2)  The  weight  of  anode  corroded  in  a  tank  per  day. 

(3)  The  deficit  in  weight  of  silver  in  the  electrolyte  in  each 
tank  per  day. 

(4)  The  electromotive  force  contributed  to  the  circuit  by  the 
solution  of  impurities  in  the  case  of  each  bullion. 

(5)  The  total  drop  of  potential  across  each  tank,  adding  10 
per  cent  for  loss  in  contacts. 

(6)  The  horse  power-hours   required  per   kilogram  of   silver 
deposited  in  each  case. 

Solution  : 

(1)   Silver  deposited  theoretically  by  one  ampere,  per  day 

0.00001035X108X60X60X24  =   96.58  grms. 
220  amperes  deposit  96.58X220  =21,248     " 

=  21.248  kg. 

19  85 
Ampere  efficiency  =  0.934  =  93.4  per  cent.  (1) 


This  means  that  6.6  per  cent  of  all  the  silver  which  the  cur- 
rent is  capable  of  depositing  is  prevented  from  depositing  by 
the  nitric  acid  present,  forming  AgNO3.  The  electrolyte  is 
silver  nitrate  with  copper  nitrate,  and  contains  always  about 
1  per  cent  of  free  nitric  acid,  which  thus  acts  chemically  upon 
the  deposited  silver  (or  acts  to  prevent  its  deposition  to  this 
extent,  whichever  way  we  wish  to  look  at  it). 

(2)  The  weight  dissolved,  per  100  grams  of  anode  corroded, 
is,  assuming  all  the  gold,  to  appear  in  the  slimes: 

No.  1. 

Anode.  Slimes.  Dissolved. 

Ag  .............  .........     98.69  0.03  98.66 

Pb  ......................        1.09  ____  1.09 

Cu  ............  .  ........  .       0.12  ____  0.12 

Fe  ......................       0.09  .....  0.09 

Au  ........  ..............       0.01  0.01 

No.  2. 

Anode.  Slimes.  Dissolved. 
Ag  .................  .....     74.08         0.14  77.94 

Pb  ......................       3.71         0.01  3.70 


608  METALLURGICAL  CALCULATIONS. 

Anode.  Slimes.  Dissolved. 

Cu 20.23  0.04  20.19 

Fe.. 1.01  0.01  1.00 

Au 0.05  0.05  .... 

The  amount  of  current  necessary  to  dissolve  these  weights 
will  be  the  sum  of  the  current  which  would  be  necessary  to 
dissolve  each  constituent  separately.  These  will  be  found  in 
ampere-hours  by  dividing  each  by  its  electrochemical  chemical 
equivalent  X  3,600. 

No.  1. 

Ampere-Hours. 

Ag 98.66^-0.001118X60X60   =   24.51 

Pb 1. 09 -5- 0.00107   X60X60   =     0.28 

Cu 0.12 -0.00033   X60X60   =     0.12 

Fe..  .   0.09-hO. 00029  X60X60   =     0.09 


25.00 

No.  2. 
Ag  ..............  77.94-4.025 

Pb  ..............   3.70^3.856 

Cu  ..............  20.19-hl.185 

Fe  ..............    1.00^-1.044 

38.32 

Since  there  is  disposable  220X24  —5,280  ampere-hours  in 
a  tank  per  day,  there  will  be  corroded  the  following  weights 
of  anodes  per  day  in  tanks  containing  Nos.  1  and  2,  respectively; 


No.  1  .....  100  x-   =  21,120  Grams. 


No.  2                                100  X  =  13,779      « 

(3)  There  are  dissolved  in  a  tank,  per  day,  the  following  weights 
of  silver: 

No.  1  anodes:  21,120X0.9866  =  20,837  Grams 

No.  2  anodes:  13,779X0.7794  =  10,739      " 

Deposit  in  each  tank  =  19,850      " 

Surplus  in  No.  1  tank  =       987      " 

Deficit  in  No.  2  tank  =    9,111      "                     (3) 


THE  METALLURGY  OF  SILVER  AND  GOLD.  609 

(4)  The  heats  of  formation  of  the  acid  and  salts  concerned 
are: 

(Ag,  N,  O3,  Ag)  =  23,000  cal.  =        213  cal.  per  gram  Ag. 

(Pb,  N2,  O«,  Ag)  =  98,200     "  =        474     "  *      Pb. 

(Cu,  N2,  O8,  Ag)  =  81,300     "  =     1,278     "  "      Cu. 

(Fe,  N2,  O6,  Ag)  =  43,900     "  =       784     "  "      Fe. 

(H,  N,  O3,  Ag)  =  48,800     "  =  48,800     "  "      H 

The  heat  balance  per  100  grams  of  anodes  No.  1  is: 

Solution  of  Ag,  98.66  X      213  =  21,015  cal. 

Pb,  1.09  X      474   =        517  " 

Cu,  0.12X    1,278  =        153  " 

Fe,   0.09  X       784  =          71  " 

Heat  evolved  21,756  cal. 

Deposition  of  Ag,  93.99  X   213=20,020  " 

Liberation  of  H2  0.06X48,800  =    2,928  " 


Heat  absorbed  =  22,948 


Heat  deficit  1,192    " 

Since  this  is  per  25  ampere-hours  [see  (2)]  and  1  ampere 
hour  at  1  volt  =  860  calories,  the  voltage  which  this  deficit  of 
energy  will  absorb  in  the  case  of  anodes  No.  1  is: 

1,192^-25^-860  =  0.06  volt.  (4) 

The  similar  calculation  for  anodes  No.  2  gives  per  100  grams 
corroded : 

Solution  of  Ag,  77.94  X    213  =  16,601  cal. 

Pb,     3.70  X    474  =     1,754    " 

."  Cu,  20.19X1,278  =25,803    " 

Fe,      1.00  X    784  =        784    " 

Heat  evolved  44,942  cal. 

Deposition  of  Ag,  98.11  X    213  =  20,897    " 
Liberation  of  H2    0.064x8,800=    2,928    " 


Heat  absorbed  23,826 


Surplus  heat  21,117    " 

Voltage  generated  =  0.98  volt.  (4) 


610  METALLURGICAL  CALCULATIONS. 

(5)  Resistance  of  each  element  of  electrolyte  of  1  square 
cm.  area  is  20X4  =  80  ohms.  The  current  passing  through 
this  is  220 -f- 10,000  =  0.022  amperes.  The  drop  of  voltage, 
due  to  the  resistance  of  the  electrolyte,  is,  therefore, 

0.022X80  =  1.76  volts. 
Adding  decomposition  voltage  we  have 

In  case  of  No.  1  anodes,  1.76  +  0.06  =-  1.82  volts. 

Add  for  resistance  of  contacts  0.18      " 


Working  voltage  2.00      "         (5) 

In  case  of  No.  2  anodes,  1.76  -  0.98  =  0.78      " 

Add  for  resistance  of  contacts  0.18      " 

Working  voltage  0.96      "         (5) 

(6)  Horsepower  required  per  tank: 

220X2.00 

Anodes  No.  1  :     —^7;  -  =  0.59  hp. 

750 


Anodes  No.  2: 

Power  per  kilogram  of  silver  deposited: 
Anodes  No.  1,  0.59-7-19.85  =  0.020  hp-days 

=  0.48  hp-hours.    (6) 
Anodes  No.  2,  0.28^19.85  =  0.014  hp-days. 

0.34  hp-hours.     6) 

Problem  128. 

The  Wohlwill  process  for  refining  gold  bullion  operates  with 
AuCl3  solution,  strongly  acid  with  HC1.  Design  a  plant  for 
refining  bullion  having  the  analysis: 

Au  ..............  60.3  Fe  ...............  2.2 

Ag  .....  :.•.'.'.  .....   7.0  Ni  .............  ..2.0 

Cu  ..............   6.5  Pb  ...............  7.0 

Zn  ..............  15.0 

Assume  the  following  data  to  start  with: 

Current  density,  1,000  amperes  per  square  meter. 

Use  ten  tanks  in  series. 

Tanks  available,  500  mm.  X  500  mm.  X300  mm.  deep,  inside, 


THE  METALLURGY  OF  SILVER  AND  GOLD.  611 

Electrodes  about  4  cm.  apart. 

Starting  sheets  1  mm.  thick. 

Anodes  20  mm.  thick. 

Specific  gravity  of  anodes  17.5. 

Gold  present  in  solution  50  grams  per  liter. 

Specific  resistance  of  electrolyte  5  ohms. 

Specific  gravity  of  the  electrolyte  1.15. 

Loss  of  electromotive  force  at  contacts  one-half  the  loss  due 
to  the  resistance  of  the  solution. 

Anode  products,  AuCl3,  AgCl,  CuCl,  ZnCP,  Fed2,  NiCl2, 
PbCl2. 

Required: 

(1)  The  size  and  dimensions  of  electrodes  and  number  in  a 
tank.    . 

(2)  Current  used  and  total  voltage  of  the  system. 

(3)  Gold  deposited  per  day. 

(4)  Anodes  corroded  per  day,  assuming  corrosion  uniform. 

(5)  Deficit  of  gold  in  the  tanks  per  day. 

(6)  If  anodes  are  left  in  until  9/10  dissolved,  how  long  will 
they  last? 

(7)  If  cathodes  are  removed  every  24  hours,  what  is  the 
average  time  of  treatment? 

(8)  At  6  per  cent  per  annum,  what  is  the  interest  charge 
per  kilogram  of  gold  under  treatment  in  the  plant,  gold  being 
worth  72.9  cents  per  gram. 

Solution : 

(1)  Allowing  1  cm.  clearance  at  the  sides,  the  plates  must 
not  be  over  48  cm.  in  width  across  the  tank.  The  electrolyte 
must  not  be  nearer  than  2  cm.  to  the  top  of  the  tank,  and  the 
plates  ought  to  be  8  centimeters  above  the  bottom,  to  allow 
space  for  slimes  to  accumulate.  The  immersed  depth  of  plates 
is,  therefore,  40  cm.,  and  the  superficial  area  960  sq.  cm.  on  a 
side. 

With  spaces  4  cm.,  anodes  2  cm.  thick  and  cathodes  0.1  cm., 
using  one  more  cathode  than  anode,  we  have  the  spaces  equal 
in  number  to  twice  the  anodes,  and,  therefore, 

(2  X  anodes)  -f  (anodes  +  1)  0.1  + (2  anodes +  4)  =  50,  whence  the 
number  of  anodes  figure  out  4.9.  (1) 


612  METALLURGICAL  CALCULATIONS. 

Choosing  the  nearest  whole  number,  5,  there  will  be  6  cath- 
odes, and  the  working  spaces  will  be  10,  and  their  width 

50  -5CO-  6(0.1)  ^94cm  (1) 

(2)  Ignoring  the  edges  of  the  plates,  the  working  surface 
per  tank  is  960X2X5  =  9,600  sq.  cm.,  and  the  current  9,600 -r- 
10,000X1,000  =  960  amperes.  (2) 

The  voltage  to  overcome  ohmic  resistance  is 

- *-97  volts- 

Add  50  per  cent  loss  at  contacts  =   0 . 98      " 


Sum  =   2.95 

The  voltage  corresponding  to  the  chemical  action  occurring 
can  be  calculated  from  the  energy  of  formation  of  the  AuCl3 
decomposed  minus  the  energy  of  formation  of  the  salts  formed. 
More  simply,  although  not  so  logically,  it  may  be  derived  from 
the  voltages  of  decomposition  of  these  compounds,  allowing  for 
the  proportions  of  each  formed  and  decomposed.  These  are: 

(Au,  Cl3)  =    27,200-^-3-^23,040  =  0.393  volt. 

(Ag,  Cl)  =    29,000^1^23,040  =1.257  '" 

(Cu,  Cl)  =    35,400^-1-^-23,040  =1.536  " 

(Zn,  Cl2)  =  113,000-^-2^-23,040  =2.448  " 

(Fe,  Cl2)  =  100,100-^2-^23,040  =2.170  * 

(Ni,  Cl2)  =     93,900  -i-  2  ~-  23,040  =2.037  a 

(Pb,  Cl2)  =    77,900 -r- 2 -f- 23,040  =1.690  « 

These  are  the  voltages  which  would  be  generated  at  the 
anode  in  case  each  one  of  these  salts  was  the  only  salt  being 
formed.  But,  since  we  know  the  exact  composition  of  our 
anode  dissolved,  we  can  calculate  the  corresponding  voltage 
generated  at  the  anode: 

Au,  0.603X0.393  =  0.237  volt. 

Ag,  0.070X1.257  =  0.088     " 

Cu,  0.065X1.536  =  0.101     " 

Zn,  0.150X2.448  =  0.367     * 

Fe,  0.022X2.170  =  0.048     « 


THE  METALLURGY  OF  SILVER  AND  GOLD.  613 

Ni,    0.020X2.037  =  0.041  volt. 
Pb,   0.070X1.690  =  0.118     " 


Sum        1.000 
Absorbed  at  the  cathode  0.393     " 

Total  voltage  generated  0.607     " 

Absorbed  in  electrolyte  and 

connections  2.95       " 

Working  voltage  per  tank        2.34       " 

Total  for  the  system  23.4         "  (2) 

(3)  Au  deposited  by  960  amperes  in  ten  tanks  per  day: 

1Q7 
0. 00001035  X^  X 60 X 60X24X960X10    =  563,729  gr. 

=  563.729  kg. 

(4)  One  gram  of  anode  requires  for  its  corrosion: 
Au,       0.6034-197,  4-  34-0.00001035 

Ag,  0.603 -=-108  4-0.00001035 

Cu,  0.065-:-   63.6       4-0.00001035 

Zn,  0.150^-   65     4-  2  4-  0.0000 1035 

Fe,  0.022  4-   56     4-24-0.00001035 

Ni,  0.020^-   59     4-2-0.00001035 

Pb,  0.070  4-207      -f-24-  0.00001035 


=  0.01703  4-0.00001035  =  1,645  amp.  seconds. 

Current  available  per  day: 
960  X  60  X  60  X  24  X 10      =  829,440,000  amp.  seconds. 

Anodes  corroded  per  day: 
829,440,0004-1,645  504,220  grams. 

604.220  kilograms    (4) 

(5)  Gold  deposited  563.729 

Gold  dissolved  =  504.220X0.603   =         304.045 


Deficit  of  gold  in  the  plant  per  day  =         259.684          u  (5) 

(6)  Weight  of  anodes: 
960X4X17.5X5X10  =      3,360,000  grams. 

3,360  kilograms. 
One-tenth  scrap  336          " 

Weight  corroded  =  3,024 


614  METALLURGICAL  CALCULATIONS. 

Days  to  corrode  them: 

o  094 

|=  5.93  =  practically  6  days.  (6) 

(7)  Average  time  of  treatment,  removing  one-sixth  each  day: 

(1  +  2  +  3  +  4  +  5  +  6)  -f-6  =  3.5  days.  (7) 

(8)  Value  of  gold  in  plant  is,  in  regular  running,  value  of 
cathodes  for  all  the  time  plus  value  of  anodes  for  3.5-f-6  =  58 
per  cent  of  the  time. 

Value  of  cathodes : 
960X0.1X19.2  (sp.  gr.  gold)  X6X  10X0.729  =      $80,578 

Value  of  gold  in  anodes : 

3,360,000X0.603X0.729  =  1,477,012 

Average  value  of  anode  gold  under  treatment: 

1,477,012  X  (3.5  -5-6)  =      861,590 

Sum  of  cathode  gold  continually  in  stock  and  average  value 
of  anode  gold  under  treatment: 

$80,578 +  $861,590   =  $942,168 

Interest  per  annum  at  6  per  cent   =      56,530 

Interest  charge  per  day  =  155 

Interest  charge  per  kg.  of  anode  refined: 

$155  -^  504 . 22  =        $0.31  per  kg. 

Interest  charge  per  kg.  of  gold  under  treatment: 

$155  -^  304 . 045  =        $0.51  per  kg.        (8) 

THE  VOLATILIZATION  OF  SILVER  AND  GOLD. 
It  is  known  that  both  of  these  metals  can  be  vaporized;  it 
can  easily  be  done  in  the  electric  arc.  If  they  are  placed  in 
a  vacuum,  in  quartz  vessels,  they  show  signs  of  metal  vapor- 
izing at  680°  and  1,070°,  respectively,  although  what  vapor 
tension  this  corresponds  to  we  do  not  exactly  know,  but  we 
can  surmise  it  to  be  a  small  fraction  of  a  millimeter.  If  heated, 
still  higher  in  a  vacuum,  they  show  the  phenomenon  of  ebul- 
lition, or  boil,  at  1,360°  and  1,800°,  respectively,  although 
here  again  we  do  not  know  what  pressure  this  corresponds  to, 


THE  METALLURGY  OF  SILVER  AND  GOLD  615 

except  it  is  the  pressure  of  the  melted  metals  above  the  spot 
where  the  vapor  commences  to  form  beneath  the  surface, 
which  might  be  5  to  10  millimeters  of  melted  metal,  say  10 
millimeters  of  mercury,  if  the  bath  of  metal  were  shallow. 

From  these  data,  obtained  by  Schuller,  Krafft  and  Bergfeld,  it 
is  estimated  that  the  boiling  points  of  the  two  metals  under 
atmospheric  pressure  are  2,040°  and  2,530°,  respectively,  on 
the  assumption  that  the  temperature  interval  between  first 
signs  of  vaporization  in  a  vacuum  and  boiling  in  a  vacuum  is 
equal  to  the  interval  between  the  latter  temperature  and  the 
ordinary  boiling  point.  (This  is  said  to  be  true  of  mercury.) 

O.  P.  Watts  has  estimated  the  boiling  point  of  silver  and 
gold  as  1,850°  and  2,800°,  respectively,  meaning  probably  at 
atmospheric  pressure;  but  his  estimate  is  based  on  such  as- 
sumptions that  they  cannot  lay  claim  to  as  much  accuracy 
as  the  data  and  estimates  above  cited. 

If  we  go  back  to  the  very  probable  rule,  that  at  equal  frac- 
tions of  the  normal  boiling  points,  expressed  in  absolute  tem- 
peratures, metals  have  the  same  vapor  tensions,  we  can  com- 
pare silver  and  gold  with  mercury,  whose  vapor  tension  is 
known  through  a  wide  range,  and  get  quite  probable  values 
for  the  vapor  tensions  of  these  metals  through  a  large  range 
of  temperature.  The  following  table  is  from  nearly  zero  ten- 
sion to  about  the  temperature  at  which  ebullition  is  noticeable 
in  a  vacuum: 


Tension 

of  Vapor. 

Mercury. 

Lead. 

Silver. 

Gold. 

mm.  of  Hg. 

C° 

C° 

C° 

C° 

0.0002 

0 

625 

729 

942 

0.0005 

10 

658 

766 

987 

0.0013 

20 

691 

802 

1,031 

0.0029 

30 

724 

839 

1,075 

0.0063 

40 

757 

876 

1,120 

0.013 

50 

790 

913 

1,165 

0.026 

60 

822 

949 

1,209 

0.050 

70 

855 

986 

1,254 

0.093 

80 

888 

1,023 

1,298 

0.165 

90 

921 

1,059 

1,343 

0.285 

100 

954 

1,096 

1,387 

616 


METALLURGICAL  CALCULATIONS. 


Tension 

of  Vapor. 

Mercury. 

Lead 

Silver. 

Gold. 

mm.  of  Hg. 

C°   ' 

C° 

C° 

C° 

0.478 

110 

987 

1,133 

1,432 

0.779 

120 

1,020 

1,169 

1,476 

1.24 

130 

1,053 

1,206 

1,520 

1.93 

140 

1,086 

1,243 

1,565 

2.93 

150 

1,119 

1,280 

1,611 

4.38 

160 

1,151 

1,316 

1,654 

6.41 

170 

1,184 

1,353 

1,699 

9.23 

180 

1,217 

1,390 

1,743 

Inspecting  the  above  table,  we  see  that  apparently  about 
0.0002  mm.  of  mercury  tension  is  sufficient  to  make  a  metal 
show  signs  of  vaporization.  The  corresponding  temperatures 
at  which  silver  and  gold  show  0.0002  mm.  tension  are,  from 
the  above  table,  729°  and  942°,  respectively,  whereas  it  is  said 
to  have  been  observed  for  these  metals  at  680°  and  1,070°. 
The  divergence  is  not  wide,  considering  the  lack  of  exact  data 
involved. 

Mercury  is  said  to  show  ebullition  in  a  vacuum  at  180°,  lead 
at  1,250°,  silver  at  1,360°  and  gold  at  1,800°.  From  the  above 
table  we  see  these  metals  having  the  same  vapor  tension  at 
180°,  1,217°,  1,390°,  and  1,743°,  respectively.  The  agreement 
is  encouraging. 

The  table  in  continuation  is  for  temperatures  from  those 
causing  ebullition  in  a  vacuum  to  those  required  to  boil  the 
metals  at  atmospheric  pressure: 


Tension 

of  Vapor. 

Mercury.         1 

^>ead. 

Silver. 

Gold. 

mm.  of  Hg. 

C° 

c° 

C° 

C° 

9.23 

180              ! 

1,217 

1,390 

1,743 

14.84 

190 

1,250 

1,427 

1,788 

19.90 

200 

1,283 

1,463 

1,832 

26.35 

210 

,316 

1,500 

1,877 

34.70 

220 

,349 

1,537 

1,921 

45.35 

230 

,382 

1,574 

1,965 

58.82 

240 

,415 

1,610 

2,010 

75.75 

250 

,448 

1,647 

2,055 

96.73 

260 

,480 

1,684 

2,099 

THE  METALLURGY  OF  SILVER  AND  GOLD.  617 


Tension 

of  Vapor. 

Mercurv. 

Lead. 

Silver. 

Gold. 

mm.  of  Hg. 

C°  ' 

C° 

C° 

C° 

123. 

270 

1,513 

1,720 

2,144 

155. 

280 

1,546 

1,757 

2,188 

195. 

290 

1,579 

1,794 

2,233 

242. 

300 

1,612 

1,830 

2,277 

300. 

310 

1,645 

1,867 

2,322 

369. 

320 

1,678 

1,904 

2,366 

451. 

330 

1,711 

1,941 

2,410 

548. 

340 

1,744 

1,977 

2,455 

663. 

350 

1,777 

2,014 

2,500 

760. 

357 

1,800 

2,040 

2,530 

For  tensions  of 

metallic  vapors  over  1 

atmosphere, 

such  as 

may  easily  occur  in   high   temperature 

work,   particularly  in 

electric  furnaces, 

the  following 

data  may 

be  useful: 

Tension 

of  Vapor. 

Mercury. 

Lead. 

Silver. 

Gold. 

Atmospheres. 

C° 

C° 

C° 

C° 

1.0 

357 

1,800 

2,040 

2,530 

2.1 

400 

1,951 

2,197 

2,722 

4.25 

450 

2,116 

2,380 

2,945 

8. 

500 

2,280 

2,564 

3,167 

13.8 

550 

2,445 

2,747 

3,390 

22.3 

600 

2,609 

2,931 

3,612 

34.0 

650 

2,774 

3,114 

3,835 

50. 

700 

2,938 

3,298 

4,057 

72. 

750 

3,103 

3,481 

4,280 

102. 

800 

3,267 

3,665 

4,502 

137.5 

850' 

3,436 

3,848 

4,725 

162. 

880 

3,525 

3,958 

4,858 

The  last  table  may  not  be  of  much  immediate  practical 
value,  but  it  is  interesting  scientific  information.  The  pre- 
ceding tables  are,  however,  technically  and  commercially 
important.  They  point  particularly  to  the  wastefulness  of 
melting  silver  or  gold  in  open  furnaces  where  furnace  gases 
pass  over  the  metal.  Such  gases,  at  temperatures  above  700° 
for  silver  and  950°  for  gold,  i.e.,  even  passing  over  unmelted 
metal,  can  cause  volatilization  and  loss  of  weight,  because  they 


618  METALLURGICAL  CALCULATIONS. 

evaporate  the  metal  on  exactly  the  same  principle  that  a  current 
of  dry  air  evaporates  ice  or  water.  They  also  explain  why 
silver  volatilizes  with  lead  in  the  cupellation  operation.  At 
1,000°  C.  lead  has  a  vapor  tension  of  about  0.62  mm.  of  mer- 
cury, and  silver  about  0.07  mm.,  and  therefore,  a  considerable 
loss  of  silver  is  sure  to  occur  with  the  lead  vapors  passing  off. 
Gold  at  the  same  temperature  has  only  0.0007  mm.  tension,  and, 
therefore,  proportionately  less  of  it  is  lost,  say  only  one-fiftieth 
as  much  as  the  silver  loss,  reckoning  from  the  proportionate 
tensions  and  the  relative  densities  of  the  two  vapors. 

In  calculating  such  metallic  vapor  losses  it  is  important  to 
remember  that  the  metals  are  monatomic  when  in  the  state 
of  vapor,  and-  that  the  molecular  weights  of  their  vapors  are 
simply  their  atomic  weights.  The  hypothetical  weight  of  a 
cubic  meter  of  such  metallic  vapor  at  standard  conditions,  0° 
C.  and  760  mm.  pressure,  is  therefore, 

0.09kg.xat°miC9Weight. 


and  from  this  theoretical  datum  the  weight  of  any  volume  at 
any  temperature  and  any  tension  can  be  calculated. 

Illustration:  Calculate  the  weight  of  silver  vapor  contained 
in  1  cubic  meter  of  furnace  gases,  if  saturated  with  silver  vapor 
and  at  1,100°  C. 

Solution:  Since  the  tension  of  silver  at  this  temperature 
is  0.28  mm.,  "the  question  resolves  itself  into  this:  What  is 
the  weight  of  1  cubic  meter  of  silver  vapor  at  1,100°  and  0.28 
mm.  pressure.  One  cubic  meter  at  standard  conditions  would 
be  0.09  X  (108  -5-2)  =  4.86  kilograms;  therefore,  at  these  given 
conditions,  it  contains: 


=  3.5  grams. 

It  is  true  that  the  silver  vapor,  being  heavy,  mixes  slowly 
with  the  furnace  gases,  but,  nevertheless,  the  gases  rushing 
over  and  coming  in  contact  with  the  silver  may  become  nearly 
saturated  with  this  vapor,  and  carry  considerable  away.  In 


THE  METALLURGY  OF  SILVER  AND  GOLD.  619 

the  Bessemerizing  of  copper  matte  to  blister  copper,  by  blowing 
air  directly  through  it,  as  much  as  30  per  cent  of  all  the  silver 
present  may  be  thus  vaporized  from  the  bath  as  soon  as  the 
silver  has  taken  the  metallic  form. 

Every  smelter  and  refiner  of  the  precious  metals  should  be 
familiar  with  these  facts,  and  draw  conclusions  from  them 
useful  to  the  conduct  of  his  business. 

The  specific  heat  of  these  metallic  vapors  is  0.225  per  cubic 
meter  (calculated  to  standard  conditions),  and  the  latent  heat 
of  vaporization  is  approximately  twenty-three  times  the  normal 
boiling  point  expressed  in  absolute  temperature  for  one  atomic 
weight  of  the  metal — as  explained  in  Part  I  of  these  calculations. 

For  silver  and  gold  we  would  have  the  latent  heats: 

Ag,  23  (2,040  +  273)   =  53,200  Cal.  per  108  kg. 

=       493        "        I  kg. 
Au,  23  (2,530  +  273)   =  64,470        "        197  kg. 

=       327        "        1  kg. 

The  above  latent  heats  of  vaporization  are  for  boiling  at 
normal  atmospheric  pressure ;  for  vaporization  at  other  pressures 
and  corresponding  temperatures,  correction  must  be  made  for 
the  difference  in  specific  heats  of  the  metallic  vapor  and  liquid 
metal. 


CHAPTER  IV. 

THE  METALLURGY  OF  ZINC. 
(INCLUDING  CADMIUM  AND  MERCURY.) 

At  the  present  time,  the  metallurgy  of  zinc  is  briefly  com- 
prehended in  the  following  statements:  The  chief  ore  is  zinc 
sulphide,  ZnS,  infusible  at  ordinary  furnace  heats,  non-volatile, 
easily  roasted  to  ZnO ; '  the  roasting  is  done  principally  in  me- 
chanically stirred  furnaces,  the  ore  being  in  small  pieces,  because 
it  is  non-porous,  compact,  and  roasts  slowly;  the  roasted  ore, 
principally  ZnO,  is  mixed  with  an  excess  of  carbon  as  a  re- 
ducing agent,  and  heated  in  closed  fire-clay  retorts  having 
condensers  attached;  zinc  vapors  begin  to  come  off  at  1033°  C., 
and  come  off  rapidly  at  the  working  temperature  of  the  charge, 
say  1200°  to  1300°;  the  zinc  vapor  and  carbon  monoxide  pass 
into  the  condensers,  and  as  they  cool  deposit  the  zinc,  some 
in  the  form  of  fine  dust  (like  hoar  frost) ,  most  of  it  as  liquid 
drops;  the  cadmium  in  the  ore  and  some  lead,  if  present,  distil 
over  with  the  zinc,  constituting  its  chief  impurities.  Arsenic 
is  sometimes  present  in  the  condensed  product.  Iron  does 
not  distil  over,  but  some  is  absorbed  from  ladles  and  moulds 
in  which  the  liquid  zinc  may  be  handled  and  cast. 

The  chief  operations  with  which  calculations  may  be  con- 
cerned are  the  roasting,  reduction  by  carbon,  condensation  of 
the  vapors,  possibility  of  blast-furnace  extraction,  of  electric 
furnace  reduction,  electrolytic  extraction,  electrolytic  refining. 

Roasting  of  Sphalerite. 

Before  roasting,  the  crude  ore  is  crushed  and  concentrated. 
Pure  ZnS  contains  67  per  cent  zinc  and  33  per  cent  sulphur; 
its  specific  gravity  is  3.9  and  it  has  fine  cleavage,  so  that  it 
crushes  easily,  producing  much  fines  or  slimes.  In  the  Joplin 
district,  in  Missouri,  the  largest  zinc  field  in  America,  the  ore 
as  mined  averages  some  4.3  per  cent  of  zinc,  equal  to  6.4  per 
cent  of  pure  blende,  ZnS,  and  is  concentrated  by  jigging  to 

620 


THE  METALLURGY  OF  ZINC.  621 

heads  carrying  about  60  per  cent  of  zinc  (90  per  cent  of  ZnS), 
containing  some  70  per  cent  of  the  zinc  in  the  ore,  and  tails 
carrying  about  1.35  per  cent  of  zinc  (2  per  cent  of  ZnS),  repre- 
senting 30  per  cent  of  the  zinc  in  the  ore.  The  concentrates 
average  5  per  cent  of  the  weight  of  the  ore;  concentration  ratio 
20  to  1.  Cost  of  such  crushing  and  concentration  20  to  40 
cents  per  ton  of  ore  milled.  (Ingalls.)  The  average  com- 
position of  these  concentrates  is: 

Zinc 60  per  cent    =  90  per  cent  ZnS. 

Iron 2         "          =    3  per  cent  FeS. 

Silica 7 

Sulphur 31 

100 

Zinc  sulphide  begins  to  be  oxidized  by  air  at  a  dull  red  heat, 
say,  600°  C.,  and  if  the  supply  of  air  is  kept  up  the  oxidation 
is  rapid,  generating  a  large  amount  of  heat  and  consequent 
high  temperature. 

2ZnS  +  3O2  =  2ZnO  +  2SO2 

The  question  as  to  how  high  a  temperature  would  theoret- 
ically result  if  sphalerite  were,  thus  burned  is  an  interesting  one. 
Ingalls  quotes  Hollaway  as  giving  1992°  C.  We  will  investigate 
this  point  as  being  of  interest  and  value  in  connection  with 
practical  roasting. 

Problem  129. 

Pure  ZnS  is  oxidized  by  air.  The  heats  of  formation  con- 
cerned are: 

(Zn,  S)  =  43,000 
(Zn,  O)  =  84,000 
(S,  O2)  =  69,260 

The  specific  heats  of  the  materials  concerned  are: 

Sm  to  O° 

ZnS  =  0.120   +  0.00003t 
Zn  O  =  0.1212  +  0.0000315t 
SO2  (1  m3)  =  0.36     +0.0003t 
O2or  N2  (    "    )  =  0.303   +  0.000027t 

Assuming  that  the  blende  is  first  heated  to  600°,  to  start  the 
roasting,  and  then  the  air  supply  put  on,  the  exposed  roasting 


622  METALLURGICAL  CALCULATIONS. 

surface  being  sufficient  for  rapid  oxidation,  and  the  tempera- 
ture too  high  to  form  SO3: 

Required: 

(1)  The  theoretical  temperature  at  the  roasting  surface  at 
starting,  if  all  the  oxygen  passing  is  utilized. 

(2)  The  same,  when  the  operation  has  continued  to  its  max- 
imum temperature. 

(3)  The  same,  if  three-fourths  the  oxygen  passing  is  utilized. 

(4)  The  same,  if  half  the  oxygen  passing  is  utilized. 

(5)  The  same,  if  the  resulting  gases  contain  only  5  per  cent 
of  SO2  gas. 

Solution : 

(1)  The  reaction  gives,  thermally: 

Decomposition  of  ZnS  —43,000  Cal. 

Formation  of  ZnO  +84,800     " 

"   SO2  +69,260     " 


Net  heat  evolution  111,060  Cal. 

This  would  be  concerned  in  the  oxidation  of  97  kg.  of  ZnS 
to  81  kg.  of  ZnO  and  64  kg.  of  SO2  (=  22.22  m3),  and  requiring 
3X16  =  48  kg.  of  O2  =  208  kg.  of  air  (containing  160  kg.  = 
127  cubic  meters  of  N2). 

Heat  in  97  kg.  of  ZnS  at  600°    =      8,032  Cal. 
Total  heat  in  the  products          =  119,092     " 

Mean  heat  capacity  of  the  products  per  1°,  from  0°  to  t°: 

81kg.  ZnO  =  9.8172 +  0.002552t 
22.22m3  SO2  =  8.0000 +  0.006667t 
127  m3  N2  =  38.4810  +  0.003429t. 


Sum  =  56.2982 +  0.012648t 

Therefore,  theoretical  surface  temperature,  at  starting, 

119,092  _  1KftRor  m 

=  56.2982 +  0.012648t  = 

(2)  When  the  surface  of  the  oxidizing  material  has  attained 
its  maximum  temperature,  it  is  practically  at  t°,  and  the  pro- 
ducts of  combustion  will  contain  111,060  Calories  plus  the 
heat  in  97  kg.  of  ZnS  at  t°.  We  then  have 


THE  METALLURGY  OF  ZINC.  623 

111,060  +  11.24t  +  0.00291t2 


56.2982  +  0.012648t 


1780°.  (2) 


It  must  be  emphasized  that  this  is  the  theoretical  maximum 
under  the  most  favorable  conditions  as  to  oxidation  and  exact 
air  supply,  conditions  never  realized  in  practice. 

(3)  The  excess  of  oxygen  would  be  48  x  J  =  16  kg.  =  11.11  m3. 
This  would  be  the  most  favorable  possible  proportion  for  mak- 
ing sulphuric  acid  from  the  gases,  since  it  would  just  serve  to 
oxidize  the  SO2  to  SO3  in  the  acid  chambers,  the  complete  re- 
actions being: 

Roasting  -ZnS  +  2O2  =  ZnO  +  SO2  +  O 
Acid  chambers-  SO2  +  O  =  SO3. 

The  ll.llm3  of  oxygen  corresponds  to  53.4m3  of  excess 
air,  whose  mean  heat  capacity  is  16.1903  +  0.001443t.  Adding 
this  to  the  mean  heat  capacity  of  the  products  we  have: 

111,060  +  11.24t  +  0.00291t2 
72.4885  +  0.014091t 

(4)  This  is«  more  nearly  the  practical  conditions,  and  apply- 
ing the  principles  above  explained,  we  have: 


111,060  +  11.  24t  +  0.00291t2^ 
104.8691  +  0.016977t 

(5)   If  the  gases  contain  5  per  cent  of  SO2  gas,  their  total 
volume  per  formula  weight  of  SO2  produced,  in  kilograms,  is: 

22.22-0.05  =  444.4m3 
and  the  volume  of  excess  air  they  contain  is: 

•     444.4  -  22.2  -  127  =  295.2  m3 
the  heat  capacity  of  this  excess  air,  per  average  1°  is: 

89.4456  +  0.0079701 
and 

111,060  +  11.  24t  +  0.00291t2 


145.7438 +  0.024947t 


=  731°  (5) 


It  follows  from  this  analysis  and  these  results,  that  effective 
auto-roasting  of  zinc  sulphide  is  practicable  provided  that  the 
ore  is  finely  divided,  so  as  to  expose  large  surface,  oxidize 
quickly,  and  utilize  the  oxygen  of  the  air  efficiently.  Without 
these  conditions,  auto-roasting  is  impracticable. 


624  METALLURGICAL  CALCULATIONS. 

It  ought  to  roast  satisfactorily  by  pot  roasting,  if  the  proper 
conditions  as  to  size  of  ore,  speed  of  blast,  thickness  of  pot 
walls,  etc.,  can  be  found. 

Problem  130. 

Average  blende  concentrates,  containing  90  per  cent  ZnS, 
3  per  cent  FeS,  7  per  cent  SiO2,  are  roasted  by  the  use  of  30 
per  cent  of  coal,  in  a  furnace  with  a  hearth  135  feet  effective 
length,  the  unroasted  sphalerite  remaining  being  2.6  per  cent 
of  the  roasted  material.  The  iron  is  roasted  to  Fe2O3.  The 
coal  carries  75  per  cent  of  carbon,  5  per  cent  of  hydrogen, 
8  per  cent  of  oxygen,  1  per  cent  of  sulphur  and  11  per  cent 
of  ash.  The  chimney  gases  average  2  per  cent  of  SO2  and 
escape  from  the  furnace  at  300°  C.  Charge  is  drawn  from 
the  furnace  at  1000°  C.  Furnace  roasts  40,000  pounds  per 
day.  Width  of  furnace,  12  feet;  height,  8  feet. 

Required : 

(1)  The  composition  of  the  roasted  ore. 

(2)  The  heat  generated  by  the  roasting  of  2,000  pounds  of 
the  ore. 

(3)  The  heat  generated  by  the  combustion  of  the  fuel. 

(4)  The  heat  in  the  hot  charge  as  drawn. 

(5)  The  heat  in  the  chimney  gases. 

(6)  The  heat  lost  by  radiation  and  conduction. 

,   (7)  The  heat  loss  calculated  to  pound-calories    per  square 
foot  of  outer  surface  per  minute. 

Solution : 

(1)  The  2.6  per  cent  of  ZnS  in  the  roasted  ore  is  per  cent 
of  it,  and  not  of  the  raw  ore.  The  simplest  method  of  getting 
the  composition  of  the  roasted  ore  is  to  represent  by  x  the 
quantity  of  ZnS  remain  unoxidized  per  100  of  raw  ore.  We 
then  have: 

Weight  of  ZnS  oxidized,  per  100  raw  ore  =  90  —  x 

"       "  ZnO  formed  =  (90  -  x)  X     S  -  75.2  -  0.835  x 

y  / 


Fe2O3      "       ==  3  X  =2.7 


THE  METALLURGY  OF  ZINC.  625 

Weight  of  SiO2  remaining  =7.0 

"   roasted  ore  =84.9- 0.835  x 

"       "        "      "    also  =  x- 0.026 

Therefore 

84.9- 0.835  x  =  x- 0.026 
Whence 

x  =  2.2 
Percentage  of  the  ZnS  oxidized 

nrv 09 

9Q       =  0.976  =  97.6  per  cent. 
Composition  and  weight  of  the  roasted  ore: 


ZnS 

=    2.2  = 

2.6 

ZnO 

=  73.4  = 

86.1 

Fe2O3 

=    2.7  = 

3.1 

SiO2 

=    7.0  = 

8.2 

Weight  85.3 


Zinc  contents  =  60.0  =  70.3  per  cent.  (1) 

(2)  Zinc  sulphide  oxidized  to  ZnO 

2000  X  (0  .  90  -  0  .  022)  =         1,756  Ibs. 

Heat  of  oxidation  of  97  Ibs.  of  ZnS  to 

ZnO  and  SO2  (from  Prob.  129)  =      111,060  Ib.-Cal. 
Heat  of  oxidation  of  1  Ib.  1,145   "      " 

"      «          "         "   1,756  Ibs.  =  2,010,620    "      «   (fc, 


(3)   Heat  of  combustion  of  1  Ib.  of  fuel  to  CO2  and  H2O  vapor: 
C  to  CO2  0.75X8100  6075  Ib'.-Cal. 

Available  Hydrogen 

0.05  -  0.01  =  0.04 
H  to  H2O  condensed 

0.04X34,500  =  1380    "      " 


Calorific  power  to  H2O  condensed  7455 

Water  formed 

0.05X9  =  0.45 
Latent  heat  of  condensation 

0.45X606.5  .  =  273 


Calorific  power  to  H2O  vapor  =  7182 


626  METALLURGICAL  CALCULATIONS 

Coal  used  per  2000  Ibs.  of  ore  600  Ibs. 

Calorific  power  =  600X7182  =  4,309,200  Ib.-cal. 

(3) 
Total  heat  generated  in  the  furnace 

2,010,620  +  4,309,200        =  6,319,820  "     " 
Proportion  of  total  generated  by  the  roasting 

2,010,620-^6,319,820  =  0.325  =  32.5  per  cent. 

(4)  Charge,  as  drawn,  per  2000  Ibs.  of  raw  ore: 

ZnS  44  Ib. 

ZnO  1468    " 

Fe203  54    " 

SiO2  140   " 


1706   « 
Heat  in  this,  at  1,000°  C. 

ZnS  44X0.150    =      6.6 

ZnO          1468X0.153    =  224.6 
Fe2O3  54X0.344    =     18.0 

SiO2  140X0.260    =    36.4 


285.6X1000  =  285,600  Ib.-Cal. 

(4) 
(5)  Sulphur  going  into  the  gases: 

S  from  ZnS    =1756X32/97    =  579.4  Ib. 
S     "      FeS    =      60X32/88    =    21.8    " 
S     "     coal    =    600  X    0.01     =      6.0    " 


Weight  of  S  =  607.2    ' 

"       "   O  for  SO2        =  607.2    " 


"       "   SO2  1214.4    " 

Volume  of  SO2— 

(1214.4  X 16)  -^  2.88    =      6,747  cubic  feet. 

Volume  of  chimney  gas — 

6,747-^-0.02  =  337,350      " 

Volume  of  CO2  in  it— 

(600X0.75X16)^1.98  =  3,636      "         a 

Volume  of  H2O  in  it— 

(600X0.45X16)^-0.81  =    '  5,333      "         a 

Volume  of  N2  and  excess  air      =321,634      "        a 


THE  METALLURGY  OF  ZINC.  627 

Heat  in  these  gases  at  300°  C. 

CO2  3,636X0.436  =       1,585  oz.-cal.  per  1° 

H2O  5,333X0.385   =       2,057  "      "       " 

SO2  6,747X0.450   =      3,036  "      "       " 

N2andO2  321,634X0.311   =100,028  "      "       " 


106,706    "      " 
=  6,669  Ib.-Cal. 
=  2,000,700   "          "      300°          (5) 

(6)    Summary  of  heat  distribution : 

Heat  available,  per  2,000  Ib.  ore  =  6,319,820  Ib.-Cal. 


Heat  in  hot  ore  drawn,  235,600 

'  chimney  gases,  2,000,700 

Loss  by  radiation  and  conduction,  4,083,620 


6,319,820   '  (6) 

(7)  Approximate  periphery  of  furnace : 

12  +  12  +  8  +  8  =  40  feet. 
Approximate  outer  surface,  including  base: 
40X135  =  5,400  sq.  feet. 
Ore  roasted  per  hour: 

40,000-24  =  l,6671b. 
Heat  radiated  and  conducted  away  per  hour: 

1)667  =  3>360>000  Ib.-Cal. 
Per  square  foot  outside  area,  per  hour: 

3,360,000-^5,400  =  622  Ib.-Cal. 

per  minute  =  10.6   "      "  (7) 


Reduction  of  Zinc  Oxide. 

Since  metallic  zinc  boils  under  atmospheric  pressure  at  930°  C., 
and  carbon  does  not  begin  to  reduce  zinc  oxide  until  1033°  is 
reached,  the  zinc  reduced  is  necessarily  obtained  in  the  state 
of  vapor.  To  make  the  reaction  proceed  fast,  a  temperature 
of  the  charge  inside  the  retorts  of  1100°  to  1300°  C.  at  the  end 


628  METALLURGICAL  CALCULATIONS. 

is  necessary.  A  large  amount  of  heat  is  absorbed  in  the  reac- 
tion, which  must  be  supplied  as  a  current  or  flow  of  heat  through 
the  walls  of  the  retort  in  order  to  keep  the  reaction  and  re- 
duction going.  Since  the  walls  of  the  retort  are  fire-clay,  aver- 
aging 4  centimeters  (1.6  inch)  thick,  there  must  be  a  consider- 
able difference  of  temperature  (temperature  head)  between 
the  inside  and  outside  of  the  retort  in  order  to  keep  up  the 
flow  of  heat  inward. 

THERMOCHEMICAL  CONSIDERATIONS. 

The  heat  of  formation  of  zinc  oxide,  at  ordinary  tempera- 
tures, is  84,800  Calories  for  a  molecular  weight,  81  kilos,  of 
oxide,  as  determined  by  thermochemical  experiment.  This 
means  that  cold  zinc  uniting  with  cold  oxygen,  and  the  hot 
product  cooled  down  to  the  same  starting  temperature,  results 
in  the  evolution  of  heat  stated.  Cold  carbon  uniting  with  cold 
oxygen  to  cold  carbonous  oxide,  CO,  gives  similarly  29,160 
Calories  per  molecular  weight,  28  kilograms,  of  gas  formed. 
These  facts  are  expressed  thermochemically  as 

(Zn,  O)     =  84,800 
(C,  O)        -.  29,160 

The  equation  of  reduction  becomes 

ZnO  +  C  '=  Zn  +  CO 
-  84,800  +29,160  =  -  55,640 

This  means  that  if  we  start  with  cold  zinc  oxide  and  cold 
carbon,  the  heat  absorbed  is  55,640  Calories  ending  up  with  cold 
products,  zinc  and  CO  at  ordinary  temperatures. 

What  is  actually  done  in  practice,  however,  is  to  first  heat 
the  ZnO  and  C  to  a  high  temperature,  at  least  to  1033°,  and 
then  to  supply  the  heat  of  the  reaction  at  that  temperature,  pro- 
ducing zinc  vapor  and  CO  gas,  both  at  1033°.  The  process  of 
reduction,  therefore,  resolves  itself  plainly  into  two  steps:  (1) 
heating  the  charge  up  to  the  reacting  temperature,  (2)  supply- 
ing the  latent  heat  of  the  chemical  change  at  that  temperature. 

Heating  up  the  Charge. 

The  charge  usually  contains  more  carbon  than  is  theoretically 
necessary  for  reduction  of  the  zinc  oxide,  because  some  is  con- 
sumed by  air  in  the  retort,  some  in  reducing  iron  oxides  and 


THE  METALLURGY  OF  ZINC.  629 

some  is  left  behind  unused;  it  is  cheaper  to  lose  carbon  than 
unreduced  zinc  oxide.  However,  making  the  calculation  on 
the  theoretical  amount  of  carbon  only  (anyone  can  modify  it 
for  any  excess  of  carbon  used  in  any  specific  case)  we  need  to 
know  the  heat  necessary  to  raise  these  reacting  substances  to 
the  temperature  of  reaction. 

The  heat  in  1  kilogram  of  zinc  oxide  at  various  temperatures  was 
determined  in  the  writer's  laboratory  as  0.1212t  +  0.0000315t2. 
The  heat  in  1  kilogram  of  carbon  at  temperatures  above  1000°  is 
represented  by  0.5t  —  120.  We  have  the  heat  necessary  to  raise 
our  81  kilograms  of  ZnO  and  12  kilograms  of  carbon  to  1033°  as 

ZnO:  0.1212   (1033)4-0.0000315   (1033)2X81 

=  159X81  =  12,879  Cal. 
C  :  [0.5  (1033)  -  120]  X 12  =  396X12=   4,752    " 


Sum  =  17,631 

This  quantity  represents  the  heat  per  81  of  oxide  and  12  of 
carbon,  or  93  of  charge  containing  65  of  zinc.  Per  1000  kilo- 
grams of  oxide,  containing  800  kilos  of  zinc,  it  would  be 

17,631X^5  =  217,670  Calories. 

ol 

It  should  be  noted  that  this  quantity  is  actually  proportional 
to  the  weight  of  the  charge,  ore  plus  reducing  agent,  and  inde- 
pendent of  the  amount  of  zinc  in  it.  A  ton  of  poor  ore  will 
practically  require  as  much  heat  to  bring  it  up  to  the  reaction 
temperature  as  a  ton  of  rich  ore,  so  that  these  costs  are  propor- 
tional to  the  weight  of  charge  treated  and  not  to  the  weight  of 
zinc  it  contains. 

If  the  charge  is  heated  electrically,  the  amount  of  electric 
power  needed  for  heating  up  can  be  calculated,  assuming  an 
average  loss  of  10  to  30  per  cent  of  the  total  heat  by  radiation 
and  conduction  from  the  furnace. 

Illustration : 

A  retort  contains  300  kilograms  of  charge  mixture  and  is 
heated  electrically  to  1033°,  the  reaction  temperature  at  an  effi- 
ciency of  75  per  cent,  by  an  electric  current  of  250  horse-power. 
How  long  will  the  heating-up  period  last. 


630  METALLURGICAL  CALCULATIONS. 

Solution : 

Heat  needed  in.  the  charge,  217,670X0.3  =  65,300  Calories. 
Heat  to  be  supplied  =  65,300^-0.75  =  87,070  Calories.  Power 
applied  supplies  635X250  =  158,750  Calories  per  hour.  Time 
required  87,070-r-158,750  =  0.55  hour  =  33  minutes. 

If  the  charge  is  heated  by  furnace  gases  outside  the  retort  we 
must  take  into  consideration  that  the  fire-clay  is  a  poor  con- 
ductor, that  the  rate  of  transmission  of  heat  through  the  walls 
falls  off  as  the  -charge  becomes  hot,  and  that  the  outside  sur- 
face of  the  retort  must  be  kept  well  above  1033°  in  order  to 
get  the  charge  to  that  temperature  in  a  reasonable  time.  The 
conductance  of  fire-clay  for  high  temperature  is  0.0031;  that 
is,  0.0031  gram-calories  pass  through  each  square  centimeter 
per  second,  if  one  centimeter  thick,  per  1°  C.  difference  of  tem- 
perature. When  the  charge  is  cold,  the  inner  surface  of  the 
retort  may  be  reduced  in  temperature  to  a  low  red  heat,  say, 
500°,  and  towards  the  end  of  the  heating-up  period  its  tempera- 
ture becomes  at  least  1033°,  while  the  outer  surface  is  kept 
continually  at  1200°,  let  us  say,  by  the  furnace  gases.  During 
the  heating-up  period  there  is  a  difference  of  temperature 
producing  heat  flow  of  700°,  for  a  short  time,  down  to,  say, 
200°,  which  we  may  average  up  as  300°  heat  difference.  The 
heat  conductivity  of  the  material  of  the  retort  and  the  thick- 
ness of  its  walls  have  everything  to  do  with  the  rate  at  which 
the  heat  can  get  through  and  the  charge  be  heated  up  to  the 
reaction  temperature. 

Problem  131. 

An  oval  zinc  retort  is  130  centimeters  long,  30  centimeters 
diameter  outside,  one  way,  and  15  centimeters  the  other,  and 
its  walls  are  3  centimeters  thick.  It  is  charged  with  30  kilo- 
grams of  ore  and  12  kilograms  of  reduction  material.  The 
temperatures  of  the  charge  in  the  retort  and  of  the  gases  out- 
side the  retort  were  as  follows: 

In  Retort  Outside  Difference 

At  starting.               0°C.  1067°C.  1067°C. 

In  0.5  hour                350  1067  717 

"    1.0     "                  600  1067  467 

"    1.5  hours              781  1067  286 

"   2.0      "                  814  1100  286 


THE  METALLURGY  OF  ZINC.  631 

In  Retort  Outside  Difference 

«  2.5  "  869  1100  231 

"  3:0  "  924  1110  187 

"  3.5  "  957  1155  198 

."  4.0  '"  935  1166  231 

"  4,5  "  935  1138  203 

"  5.0  "  946  1144  198 

"  5.5  "  946  1155  209 

"  6.0  «  979  1166  187 

"  6.5  •'  1001  1177  176 

"  7.0  "  1034  1177  143 

Average,  319 

Take  159  Calories  per  kg.  as  the  heat  required  to  bring  the 
ore  to  1034°  and  396  for  the  reduction  material. 

Required : 

The  average  heat  conductivity  in  C.  G.  S.  units  of  the  mate- 
rial of  the  retorts,  in  the  lange  given,  assuming  the  inner  sur- 
face of  the  retort  to  be  at  the  same  temperature  as  the  charge. 

Solution: 
Heat  passing  through  the  retort  walls  in  7  hours: 

159X30    =  4770  Calories 
396X12    =  4752 


9522 

Per  second  =  0.378       " 

=    378  gram-cal. 
Surface  of  retort: 

Periphery         V30X15X3.14  =  66  cm. 
Area  of  sides,  66X130  =  8580  sq.  cm. 

Heat  passing  through  each  square  cm.  per  second 

378  +  8580  =  0.044  cal. 
Heat  passing  per  1°  difference 

0.044-319  =  0.00014  cal. 

Since   thickness   is   actually   3   centimeters,   conductance   in 
C.  G.  S.  units  is: 

0.00014X3  =  0.00042. 


632  METALLURGICAL  CALCULATIONS. 

Correction : 

This  coefficient  of  conductance  is  far  too  low.  The  reason 
is  that  the  inner  temperature,  the  temperature  of  the  charge,  is 
always  lower  than  the  temperature  of  the  inside  surface  of  the 
retort,  and  the  difference  of  temperature  between  the  outer' and 
inner  walls  of  the  retort  must  have  been  far  less  than  the  aver- 
age, 319°.  If  we  take  the  coefficient  of  conductance  as  deter- 
mined by  experiment  for  firebrick,  viz.,  0.0031,  then  the  average 
difference  of  temperature  between  the  inner  and  the  outer 
walls  of  the  retort  would  be: 

0.00042 
•9X  0.0030  ~          C' 

This  is  more  likely,  than  that  the  conductance  of  the  retort 
material  should  be  So  extraordinarily  low.  In  fact,  we  are  led 
to  the  conclusion  that  the  poor  heat  conductivity  of  the  charge 
itself  is  the  chief  obstacle  to  its  rapid  heating,  and  that  pre- 
heating of  the  charge  would  be  very  advisable  if  it  could  be 
done  by  some  of  the  waste  heat  of  the  gases  leaving  the  furnace. 

Distillation  of  the  Charge. 

The  driving  off  of  the  zinc  is  altogether  a  different  operation 
from  the  heating  up  of  the  charge  to  the  stated  reduction  tem- 
perature. It  is  an  endothermic  chemical  operation,  comparable 
to  the  boiling  of  water  at  a  constant  temperature,  the  latent 
heat  of  the  chemical  reaction  is  exactly  comparable  to  the 
latent  heat  of  vaporization.  The  temperature  must  be  kept 
up  to  the  temperature  of  reduction  in  order  for  the  reaction 
to  take  place  at  all,  and  then  heat-calories  must  be  supplied  at 
that  temperature  to  keep  the  reaction  going,  and  the  reduction 
proceeds  pari  passu  with  the  quantity  of  heat  supplied  at  that 
constant  temperature.  The  question  now  is:  what  is  the  latent 
heat  of  this  reaction  at  the  reaction  temperature. 

We  must  for  this  purpose  know  the  heat  of  formation  of  the 
substances  involved,  ZnO  and  CO,  at  1033°.  The  method  of 
calculating  these  is  too  long  to  insert  here,  but  may  be  learned 
from  the  writer's  book  on  Metallurgical  Calculations,  Part  I, 
p.  51.  We  have  the  heats  of  formation  from  the  elements  as 
they  exist  at  1033°: 


THE  METALLURGY  OF  ZINC.  633 

(Zn,  O)1033  =  112,580 
(C,  O)1033     =     30,091 

and  the  reaction  at  1033° 

ZnO  +  C  =  CO  +  Zn 
-  112,580  +  30,091  =  -82,489 

This  is  seen  to  be  nearly  50  per  cent  greater  than  the  heat 
of  the  reaction  calculated  to  ordinary  temperatures,  from  the 
ordinary  heats  of  formation  as  taken  from  thermochemical 
tables.  The  metallurgist  should  understand  this  difference,  for 
it  is  one  of  the  utmost  importance  in  thermochemical  calcula- 
tions, if  we  want  our  calculations  to  represent  actual  conditions 
and  to  check  up  with  practice. 

This  amount  of  heat  is  proportional  to  the  zinc  oxide  re- 
duced or  to  the  zinc  distilled  from  the  charge,  but  not  to  the 
weight  of  ore  charge  itself.  This  requirement  will,  therefore, 
be  greater,  per  retort  full  of  material,  the  richer  the  charge  is 
in  zinc.  It  is  a  constant  requirement  for  a  given  output  of 
zinc,  and  not  per  given  weight  of  ore  treated. 

The  heat  required  for  the  reduction,  therefore,  as  distin- 
guished from  the  heating-up  period,  is 

oo  400 

1*       =          1,018  Calories  per  kg.  of  ZnO  reduced 

ol 

82'^89  =          1,269        "          "      "     "  Zn  distilled  off 
o5 

=  1,269,000        "          "     ton   "   " 

Furnishing  this  reduction  heat,  at  the  high  temperature  re- 
quired, is  the  larger  part  of  the  heat  needed  in  the  whole  pro- 
cess. We  see  that  it  is  some  4.7  times  the  amount  of  heat 
needed  to  raise  the  materials  to  the  reduction  temperature. 

Problem  132. 

The  retort  charge  of  Problem  131  was  kept  at  the  reduction 
temperature  for  14  hours,  the  temperature  outside  the  retort 
being  gradually  raised  to  1300°  and  the  average  difference  in 
temperature  between  the  gases  and  the  charge  being  147°,  and 
there  being  distilled  from  the  charge  in  that  time  18  kilograms 
of  zinc. 


634  METALLURGICAL  CALCULATIONS. 

Required: 

The  average  heat  conductance  in  C.  G.  S.  units  of  the  mate- 
rial of  the  retort  from  the  above  data  and  assumptions. 

Solution : 

The  heat  furnished  in  the  14  hours  was  as  follows: 

1,269X18     =  22,842  Calories. 
Heat  furnished  per  hour: 

22,842-14  =     1,632 
Heat  furnished  per  second: 

=     0.453 

=       453  calories 

Heat  passing  each  sq.  cm.  of  retort  surface  per  second: 

453 -t- 8,580  =    0.053  calories. 
Per  1°  difference: 

0.053^-147  =    0.00036    " 
Conductance,  in  C.  G.  S.  units: 

0.00036X3  =    0.00108 

Remarks:  This  conductance  calculates  out  2.5  times  as  great 
as  from  the  data  on  the  heating-up  period,  the  reason  of  the 
higher  value  being  that  the  charge  is  at  nearly  uniform  tem- 
perature during  this  reduction  period,  and  therefore  the  dif- 
ference between  the  temperature  taken  in  the'  middle  of  the 
charge  and  the  true  temperature  of  the  inner  walls  of  the  retort 
is  less  than  before,  and  the  error  from  this  source  less.  If  we 
make  the  same  assumption  as  in  the  correction  to  the  previous 
problem,  i.e.,  take  the  conductance  of  the  retort  material  as 
0.0030,  the  difference  in  temperature  of  the  outer  and  inner 
walls  of  the  retort  during  this  period  would  calculate  out 

147X0.00108 
0.0030 

while  the  center  of  the  charge  was  then  147  —  53  =  94°  cooler, 
on  an  average,  than  the  inner  wall  of  the  retort  from  which  it 
was  deriving  its  heat. 

These  figures  appear  reasonable,  and  the  writer  would  con- 


THE  METALLURGY  OF  ZINC.  635 

elude,  from  the  data  so  far  available,  that  the  conductance 
0.003  probably  represents  a  good  approximation  to  the  correct 
value  for  zinc  retort  material,  but  that,  in  order  to  use  it,  we 
should  have  more  experimental  data  as  to  the  average  difference 
between  the  temperature  of  the  inner  walls  of  the  retort  and 
the  temperature  of  the  charge  at  various  points  in  the  retort. 

Problem  133. 

A  zinc  ore  containing  50  per  cent  of  zinc  is  mixed  with  40 
per  cent  of  its  weight  of  small  anthracite  coal,  and  retorted 
in  a  Belgian  furnace.  The  recovery  of  zinc  was  82  per  cent 
of  the  zinc  content  of  the  ore.  The  consumption  of  anthracite 
to  heat  the  furnace  was  2.25  tons  per  10  of  ore.  The  anthracite 
contained  90  per  cent  of  carbon  and  10  per  cent  of  ash,  and 
had  a  calorific  power  of  7500. 

Required : 

(1)  The  total  consumption  of  fuel  per  1000  of  zinc  obtained. 

(2)  The  efficiency  of  transfer  of  heat  from  the  furnace  gases 
to  the  charge. 

Solution : 

(1)  Zinc  charged,  per  1000  of  zinc  obtained 

1 000 -H  0.82  =  1220 
Ore  charged,  per  1000  of  zinc  obtained 

1220-^0.50  =  2440 
Coal  charged  with  ore 

2440X0.40  =    976 
Coal  burned  on  grate 

2440X2.25  =  5490 
Total  coal  used,  per  1000  of  zinc  obtained 

976  +  5490    =6466  (1) 


(2)  Calorific  power  of  coal  burned 

5490X7500    =  41,170,000 
Heat  required  to  raise  ore  to  reduction  point 
2440  X   159    =       387,960 


636  METALLURGICAL  CALCULATIONS. 

Heat  required  to  raise  fuel  to  reduction  point 

Ash  98  X   159    =          15,580 
Carbon  878  X   396    =        347,690 

Total  to  raise  charge  to  reduction  point 

751,230 
Heat  absorbed  in  distilling  away  1000  of  zinc 

1,269,000 
Total  heat  utilized 

=     2,020,230 

Thermal  efficiency  of  utilization  of  the  fuel 


=  0.049  =  4.9  per  cent.  (2) 


Problem  134. 

Natural  gas  from  lola,  Kan.,  has  the  following  composition: 


CH4 

93  per  cent. 

H2 

2 

CO 

1 

C2H4 

1 

N2 

3         " 

It  is  used  in  a  zinc  retort  furnace,  at  an  efficiency  of  transfer 
of  heat  to  the  charge  of  4.9  per  cent,  as  calculated  in  Prob. 
133,  working  a  charge  which  absorbed  2,000,000  Calories  per 
1000  kilograms  of  zinc  distilled  off. 

Required:  The  volume  of  natural  gas  required  to  be  used 
to  displace  the  anthracite  fuel  used  per  1000  kilograms  of  zinc 
produced.  The  cubic  feet  of  gas  per  1000  Ibs.  of  zinc  produced. 

Solution : 

Calorific  power  of  the  gas 

CH4  0.93X  8623  =  8009 

H2  0.02X29030  =    581 

C2H4  0.01X14365  =     144 

CO  0.01  X  3062  =      31 

8765 


THE  METALLURGY  OF  ZINC.  637 

This  result  may  be  called  Calories  per  cubic  meter  of  gas,  or 
ounce-calories  (1°  C.)  per  cubic  foot  according  to  whether  it 
is  desired  to  work  in  metric  units  or  the  English  system. 

Cubic  meters  of  gas  required,  per  1000  kilograms  of  zinc 
produced: 

2,000,000 


0.049 

Per  1000  Ibs.  of  zinc  produced: 
2,000.000 


0.049 


X 16  ^8765  =  75,200  cubic  feet. 


Electric  Smelting  of  Zinc  Ores. 

In  an  interesting  communication  to  the  American  Electro- 
chemical Society  (Vol.  XII,  p.  117),  Gustave  Gin  calculates  the 
electric  power  necessary  for  the  smelting  of  several  varieties 
of  zinc  ore,  assuming  that  the  zinc  vapor  and  other  gases  pass 
out  of  the  furnace  at  the  usual  high  temperature — 1200°  C. 

Mr.  Gin  makes  his  calculations  of  heat  required  on  the  basis 
of  molecular  weight  of  zinc  compound  reduced;  that  is,  for  81 
parts  of  ZnO  and  65  parts  of  Zn.  Calculating  to  1200°,  he  finds 
the  heat  in  the  products  Zn  and  CO  to  be 

In  65  parts  Zn 26,720  Cal. 

u   28      "      CO..  .   8,280     " 


Sum 35,000  Cal. 

whereas  we  calculate  for  the  same  quantities 

In  65  parts  Zn 37,695  Cal. 

"28      "      CO ....: 8,945     " 


46,640  Cal. 

The  difference  between  these  numbers  is  principally  in  the 
latent  heat  of  vaporization  of  zinc,. which  Gin  assumes  as  15,370 
Calories,  but  which  by  a  method  of  evaluation  used  by  the 
writer  figures  out  27,670  Calories,  and  almost  exactly  the  same 
value  has  been  obtained  by  a  different  method  by  W.  McA. 
Johnson. 

Assume  then,  that  we  start  with  cold  materials  and  end  with 


638  METALLURGICAL  CALCULATIONS. 

the  products  of  the  reaction  leaving  the  retort  at  1200°,  the 
sum  total  of  usefully  applied  heat  is  the  heat  of  the  reaction 
calculated  at  ordinary  temperature  plus  the  sensible  heat  of 
the  products  at  1200°,  or 

Heat  absorbed  by  reaction,  ordinary  temperature. .  .  .84,800  Cal. 
Heat  in  necessary  products,  at  1200° 46,640 


Total 131,440 

To  this  must  be  added  the  sensible  heat  in  the  residue  left 
in  the  retort,  to  get  the  total  heat  which  has  been  applied  to 
the  charge.  If  the  charge  were  pure  ZnO,  with  the  theoretical 
amount  of  carbon,  the  residue  would  be  nil,  but  in  practice 
there  is  always  a  residue  of  gangue  with  unused  carbon. 

Mr.  Gin,  having  calculated  the  heat  requirement  on  the  above 
basis,  then  makes  his  calculations  of  power  required  per  ton 
of  ore  smelted  in  the  following  ingenious  way:  The  weight 
of  each  component  of  1000  kg.  of  ore  is  divided  by  its  mole- 
cular weight,  and  thus  the  number  of  molecular  weights  of 
material  in  one  ton  of  ore  determined,  which,  so  to  speak, 
gives  a  kind  of  chemical  formula  for  the  ore.  An  example  will 
make  this  clear. 

Composition  of  a  calcined  calamine  zinc  ore : 

ZnO  40.50  per  cent. 

ZnSiO3  38.07 

Fe2O3  9.60 

APSiO5  8.10 

CaO  2.80 

If  we  divide  the  weight  of  each  compound  present  by  its 
molecular  weight  we  get  the  relative  number  of  molecules 
present  in  the  ore,  and  the  formula  for  the  ore: 

KG.  IN 
1000  KG. 

ZnO  405  -4-  81  =  5.0  molecular  weights. 

ZnSiO3  380.7  -f-141  =  2.7 

Fe203       96.  -h 140  =  0.6 

APSiO5     81  -7-162  =  0.5  tt 

CaO          28  -h   56  =  0.1  " 


THE  METALLURGY  OF  ZINC.  639 

The  ore  may  therefore  be  represented  by  the  formula 
5ZnO  +  2.7ZnSi03  +  0.6Fe2O3  +  0.5APSiO5  +  0.  ICaO. 

Assuming  that  the  Fe203  becomes  FeSiO3,  and  that  there  is 
to  be  added  enough  CaO  to  form  CaSiO3  with  the  rest  of  the 
SiO2  of  the  zinc  silicate,  we  will  need  1.5  CaO  to  do  it,  and 
since  there  is  0.1  CaO  present,  1.4  CaO  must  be  added,  which 
represents  1.4X56  =  79  kg.  of  CaO.  To  form  CO  with  the 
oxygen  combined  with  the  zinc  and  iron,  reducing  the  latter 
to  FeO,  will  require: 

for  5     ZnO  5.0  C. 

«    2.7ZnSiO3        2.7  " 
"    0.6  Fe2O3  0.6  " 

Sum  8.3  C.  =  100  kg. 

If  we  use  twice  the  theoretical  amount  of  carbon  needed  for 
reduction  we  have  the  following  balance  sheets;  weight  in  kilo- 
grams being  enclosed: 

ORE  ADDED  CHARGE 

5  ZnO  (405)  5  ZnO  (405) 

2.7  ZnSiO3       (381)  2.7  ZnSiO3  (381) 

0.6Fe2O3         (96)  ....  0.6  Fe2O3  (96) 

O.SAPSiO5       (81)  O.SAPSiO5  (81) 

0.1  CaO  (28)  1.4  CaO  (79)  1.5  CaO  (79) 

16.6  C  (200)  16.6  C  (200) 

The  distribution  of  this  charge  will  be  as  follows: 

CHARGE.  GASES.  RESIDUE. 

5  ZnO        (405)  5  Zn      (325) 

2.7ZnSi03   (381)          2.7  Zn      (175)  

0.6Fe2O3      (96)  1.2  FeSiO3  (158) 

0.5  APSiO5  (81)  0.5  APSiO5  (81) 

1.5  CaO        (79)  1.5  CaSiO3  (174) 

16.6  C             (200)  8.3  CO     (232)            8.3  C            (100) 

The  essential  reactions  involved  in  the  reduction  are  expressed 
by  the  reaction  formula: 

5ZnO  +  2.7ZnSiO3  +  0.6Fe2O3  +  1.5CaO  +  8.3C 

.2FeSiO3+1.5CaSiO3  +  8. 


640  METALLURGICAL  CALCULATIONS 

The  heat  requirements  per  ton  of  ore  treated  may  therefore 
be  figured  out  as: 

Calories. 
Decomposition  of  7.7ZnO  =  84,800X7.7      =  652,960 

2.7ZnSiO3 
into  ZnO  and  SiO2  =  10,000X2.7       =    27,000 

Decomposition  of  0.6Fe2O3 

into  FeO  and  O       =  64,200X0.6       -    38,520 

Heat  absorbed  in  decompositions  =  718,480 

Formation  of  1.2  FeSiO3 

from  FeO  and  SiO2  =8,900X1.2  =  10,680 
Formation  of  1.5  CaSiO3 

from  CaO  and  SiO2  =  17,850X1.5  =  26,770 
Formation  of  8.3  CO  =  29,160X8.3  =222,030 

Heat  evolved  in  formation  heats  259,480 


Net  heat  of  chemical  reactions  absorbed  459,000 

Sensible  heat  in                     7.7  Zn       =  46,640X7.7  =  359,130 

8.3  CO      =    8,945X8.3  =     74,240 

"              "                       8.3  C         =    5,760X8.3  =    47,810 

11.2  FeSi'O3 
0.5  APSiO5 
1.5  CaSiO3 

=  413  kilograms  =  460X413  =  200,000 

Sensible  heat  in  products  and  residue  =  681,180 

AoM  heat  absorbed  in  chemical  reactions  =  459,000 


Total  heat  requirement  of  the  charge  =  1,140,180 

Important  principle :  The  heat  absorbed  in  an  electric  furnace 
by  chemical  reactions  is  electric  energy  utilized  at  an  efficiency 
of  100  per  cent.  It  is  only  part  of  the  sensible  heat  of  the 
materials  being  treated  which  is  lost  by  radiation  and  con- 
duction. Heat  losses  by  radiation  and  conduction  in  electric 
furnace  processes  should  be  expressed  upon  the  total  energy  of 
the  current  diminished  by  the  heat  absorbed  in  chemical  reac- 
tions, and  not  upon  the  total  energy  of  the  current.  This 
principle  has  been  expressed  most  clearly  by  Mr.  F.  T.  Snyder, 


THE  METALLURGY  OF  ZINC.  641 

of  Chicago,   the  pioneer  electric  furnace  zinc  metallurgist  of 
America. 

Expressed  thus,  electric  furnaces  give  higher  efficiencies  the 
greater  the  absorption  of  heat  in  chemical  reactions  taking 
place  within  them.  In  mere  physical  processes,  such  as  melting, 
electric  furnaces  on  a  large  scale  give  75  per  cent  net  efficiency, 
with  25  per  cent  loss  by  radiation  and  conduction.  In  the 
above  case,  figured  out  for  1000  kg.  of  zinc  ore,  40  per  cent 
of  the  total  heat  requirement  is  absorbed  as  chemical  heat  at 
100  per  cent  efficiency,  and  60  per  cent  is  needed  as  sensible 
heat.  This  60  per  cent  then  represents  the  net  sensible  heating 
effect,  and  the  loss  of  heat  by  radiation  and  conduction  must 
be  calculated  as  one-third  of  this  quantity,  and  not  one-third 
of  the  total  net  heat  requirement.  We  therefore  have  : 

Calories. 
Net  heat  requirements  for  chemical  reactions  .........    459,000 

«       «  «,  "    sensible  heat  ...............    681,180 

Loss  by  radiation  and  conduction  ...................    227,060 

Gross  heat  requirement  of  the  furnace  ...............  1,367,240 

Kilowatt  hours  of  current  required 


Mr.  Gin's  figures  have  been  modified  by  the  writer,  as  ex- 
plained above,  and  anyone  consulting  his  paper  on  this  subject 
may  make  similar  modifications  to  the  other  cases  cited  in  that 
paper. 

Mr.  F.  T.  Snyder,  of  the  Canada  Zinc  Company,  at  Van- 
couver, B.  C.,  has  operated  the  first  practical  electric  zinc 
smelting  furnace  in  America.  Treating  mixed  lead  and  zinc 
ores,  Mr.  Snyder  uses  the  furnace  and  process  protected  in  his 
United  States  patents  of  July  2,  1907.  (See  Electrochemi- 
cal and  Metallurgical  Industry,  V.  323,  489.)  The  lead  is 
obtained  liquid,  and  the  zinc  also  condensed  to  the  liquid  state 
before  leaving  the  furnace  proper,  the  heat  of  condensation 
being  partly  absorbed  by  water  cooling  the  condensers  and 
partly  by  the  descending  charges,  which  carry  it  back  into  the 
focus  of  the  furnace.  Under  these  circumstances,  Mr.  Snyder 
reports  that  he  has  attained  unexpectedly  low  results  as  to 


642  METALLURGICAL  CALCULATIONS. 

power  requirement.     It  will  be  recalled,  that  in  discussing  the 
question  of  the  electric  smelting  of  zinc  ores  in  general,  and 
Mr.  Gin's  process  in  particular,  we  assumed  the  zinc  vapor  and 
CO  gas  to  escape  from  the  furnace  at  a  minimum  of  1033°  C. 
If,  as  in  Mr.  Snyder's  furnace,  they  escape  at  about  500°,  the 
zinc  liquid,  the  heat  in  these  hot  products  is  reduced  very  con- 
siderably, particularly  that  in  the  zinc.     Mr.  Snyder  states  in 
discussing  Mr.  Gin's  paper  (loc.  cit.)  that  he  has  smelted  pure 
zinc  oxide  at  an  expenditure  of  1050  kilowatt  hours  per  1000 
kilograms  of  oxide.     Let  us  see  how  this  coincides  with  the 
theoretical  figures: 

Calories 
Heat  value  of  1050  kw-hours  ..........  .  .  =  1050X860  =  903,000 

Heat  of  the  chemical  reactions,  assumed  to  take  place 

at  ordinary  temperatures  .............  1000X687  =  687,000 

Sensible  heat  in  products,  at  500° 

Zinc:  800  X  80    =  64,000 
CO:    342X152    =  52,000 

=  116,000 
Leaving,  by  difference  for  radiation,  conduction  and 

cooling  water  =  100,000 

Mr.  Snyder  does  not  give  details  as  to  the  exact  working  of 
his  furnace,  but  his  claim  to  reduce  a  ton  of  zinc  ore  with  1000 
kw-hours  is  seen  to  be  a  possibility,  if  he  can  remove  the  zinc 
as  liquid  zinc  from  the  furnace,  not  lose  too  much  heat  in  cool- 
ing water,  and  get  most  of  the  heat  of  condensation  of  the  zinc 
vapor  usefully  returned  by  the  descending  charges  into  the 
working  focus  of  the  furnace. 

Snyder's  furnace,  working  as  claimed,  would  show  a  useful 
efficiency  of 


on  the  current  used,  with  12  per  cent  losses.  The  real  heat 
lo'sses,  however,  should  be  expressed  upon  the  energy  used  less 
that  absorbed  in  chemical  reactions,  or  as  100,000  loss  on 
216,000  used  for  sensible  heat, 


=  0.46  =  46  per  cent. 


THE  METALLURGY  OF  ZINC.  643 

This  shows  that  the  furnace  loses  by  radiation,  conduction 
and  cooling  water  46  per  cent  of  the  energy  developed  as  sen- 
sible heat  in  the  furnace,  but  the  latter  item  is  only  24  per 
cent  of  the  total  energy  applied  to  the  furnace. 

In  short,  about  three-quarters  of  all  the  energy  consumed  by 
the  furnace  is  utilized  in  the  heat  of  the  chemical  reactions 
produced;  of  the  other  one  quarter,  half  of  it  is  represented 
by  the  sensible  heat  of  the  products  leaving  the  furnace  and 
half  is  lost  by  radiation,  conduction  and  cooling  water. 


Zinc  Vapor. 

The  condensation  of  zinc  from  the  state  of  vapor  follows  the 
same  laws  as  regulate  the  condensing  of  all  gases.  If  it  is  by 
itself  in  a  cooler  space,  it  condenses  by  contact  with  the  walls 
as  finely  divided  zinc  dust  (blue  powder)  if  the  walls  are  cold, 
as  liquid  zinc  if  the  walls  are  hot,  and  it  keeps  on  condensing 
until  the  tension  of  the  remaining  vapor  is  equal  to  the  maxi- 
mum tension  of  zinc  vapor  at  the  temperature  of  the  condensing 
vessel.  This  condition  having  been  reached,  no  more  will  con- 
dense until  the  condenser  is  cooled.  At  any  temperature,  there- 
fore, an  amount  of  zinc  remains  uncondensed  corresponding  to 
the  maximum  vapor  tension  of  zinc  at  that  temperature. 

If  other  indifferent  gases  are  present,  the  zinc  vapor  sustains 
only  part  of  the  atmospheric  pressure,  or  pressure  on  the  whole 
mixture,  so  that  condensation  cannot  begin  until  a  lower  tem- 
perature is  reached;  that  is,  a  temperature  at  which  the  maxi- 
mum vapor  tension  of  the  zinc  equals  its  partial  vapor  tension 
in  the  gas  mixture.  At  this  point  condensation  begins  and  con- 
tinues as  the  temperature  falls,  the  gas  mixture  always  being 
saturated  with  zinc  vapor.  The  phenomenon  is  precisely  simi- 
lar to  the  production  of  rain  by  the  cooling  of  air  containing 
moisture. 

As  to  what  the  vapor  tension  of  zinc  is,  the  data  are  scanty. 
The  boiling,  point  under  atmospheric  pressure  is  930°  C.,  where 
its  vapor  tension  is  760  mm.  of  mercury  column.  According  to 
Barus,  the  maximum  tension  increases  6.67  mm.  for  each  1°  C. 
increase  of  temperature.  This  is  a  difficult  determination,  but 
corresponds  satisfactorily  with  the  calculated  increase  from  an- 
alogy to  water  and  mercury. 

Water,  boiling  at  100°  C.  (373°  absolute),  varies  1°  in  boiling 


644  METALLURGICAL  CALCULATIONS. 

point  for  27.20  mm.  variation  in  vapor  tension ;  mercury,  boiling 
at  357°  C.  (630°  absolute)  varies  1°  in  boiling  point  for  12.66 
mm.  variation  in  vapor  tension.  The  ratio  of  the  two  varia- 
tions is  seen  to  be  somewhere  in  the  inverse  ratio  of  the  two 
boiling  points. 

630  27.20 

373  12.66 

If  we  compare  mercury  with  zinc,  with  the  two  boiling  points 
357°  and  930°  (630°  and  1203°  absolute),  and  the  observed  va- 
riations in  vapor  tension  at  the  normal  boiling  point  of  12.66 
mm.  and  6.67  mm.  per  1°  rise  in  boiling  temperature,  the  two 
ratios  are 

1203  12.66  _ 

630   ~  "6^7" 

This  coincidence  justifies  us  completely  in  assuming  that  the 
vapor  tension  curve  of  zinc  may  be  calculated  directly  from 
that  of  mercury,  by  assigning  for  any  given  vapor  tension  an 
absolute  temperature  to  the  zinc  vapor  of  1.91  times  that  of 
mercury  vapor.  For  cadmium  we  may  use  the  similar  ratio 
of  the  normal  boiling  points  in  absolute  degrees: 

273  +  780        1053 
273  +  357         630 

and  thus  calculate  the  cadmium  curve  from  the  mercury  curve. 
From  incipient  vaporization  to  ebullition  in  a  vacuum. 

Tension 

of  vapor             Mercury        Cadmium  Zinc 

mm.  of  Hg.  C°.  C°.  C°. 

0.0002  0  183  248 

0.0005  10  200  267     . 

0.0013  20  216  286 

0.0029  30  233  305 

0.0063  40  250  324 

0.013  50  267  344 

0.026  60  283  363 

0.050  70  300  382 


THE  METALLURGY  OF  ZINC.  645 

Tension 

of  vapor  Mercury        Cadmium  Zinc 

mm.  of  Hg.  C°.  C°.  C°. 

0.093  80  317  401 

0.165  90  333  420 

0.285  100  350  439 

0.478  110  367  458 

0.779  120  383  477 

1.24  130  400  496 

1.93  140  417  516 

2.93  150  433  535 

4.38  160  450  554 

6.41  170  467  573 

9.23  180  483  592 

Before  proceeding  further,  we  would  remark  that  cadmium 
melts  at  320°  and  zinc  at  419°;  that  both  can,  therefore,  show 
vaporizing  phenomena  nearly  150°  below  their  melting  points. 
At  9  mm.  tension,  180°  C.,  mercury  begins  to  show  the  phenom- 
ena of  ebullition,  and  we  would,  therefore,  expect  cadmium 
to  "  simmer  "  at  483°  and  zinc  at  592°.  Neither  of  these  obser- 
vations has  as  yet  been  experimentally  investigated,  so  far  as 
the  author  knows. 


From  ebullition 

in  vacuum 

to  normal  boiling 

point. 

Tension 

of  vapor 

Mercury 

Cadmium 

Zinc 

mm.  of  Hg. 

C°. 

C°. 

C°. 

9.23 

180 

483 

592 

14.84 

190 

500 

611 

19.90 

200 

517 

630 

26.35 

210 

533 

649 

34.70 

220 

550 

668 

45.35 

230 

567 

687 

58.82 

240 

584 

706 

75.75 

250 

600 

726 

96.73 

260 

617 

745 

123. 

270 

634 

764 

155. 

280 

650 

783 

195. 

290 

667 

802 

242. 

300 

684 

821 

300. 

310 

700 

840 

646  METALLURGICAL  CALCULATIONS. 

Tension 

of  vapor         Mercury         Cadmium  Zinc 

mm.  of  Hg.  C°.  C°.  C°. 

369.  320  717  859 

451.  330  734  878 

548.  340  750  897 

'     663.  350  767  915 

760.  357  780  930 

The  above  table  contains  the  more  important  data  for  the 
actual  condensation  of  zinc,  cadmium  and  mercury  vapors  in 
practice.  Thus,  if  a  mixture  of  zinc  or  cadmium  vapor  with 
an  equal  volume  of  indifferent  gas  goes  into  a  condenser,  the 
partial  pressure  of  the  metallic  vapor  being  in  this  case  only 
half  an  atmosphere,  or  380  mm.,  no  metal  will  commence  to 
condense  until  the  temperature  of  the  gases  is  reduced  to  862° 
for  zinc  and  720°  for  cadmium.  If  mercury  vapor  from  a 
roaster  is  mixed  with  19  times  its  volume  of  other  gas,  so  that 
it  only  forms  5  per  cent  of  the  mixture,  its  partial  tension  will 
be  only  5  per  cent  of  760  mm.  =  38  mm.,  and  no  mercury  will 
commence  to  condense  until  the  temperature  of  the  gas  mixture 
is  reduced  to  224°.  These  temperatures  are  exactly  analogous 
to  the  phenomenon  of  the  dew  point  of  moist  air,  the  tempera- 
ture at  which  rain  forms. 

From  normal  boiling  point  to  high  pressures. 
Tension 

of  vapor  Mercury  Cadmium  Zinc 

atmospheres.  C°.  C°.  C°. 

1.0  357  780  930 

2.1  400  851  1012 
4.25  450  934  1107 
8.  500  1018  1203 

13.8  550  1101  1298 

22.3  600  1185  1394 

34.0  650  1268  1489 

50.  700  1352  1585 

72.  750  1435  1680 

102.  800  1519  1776 

137.5  850  1602  1871 

162.  880  1652  1928 


THE  METALLURGY  OF  ZINC.  647 

The  latter  table  may  have  several  special  applications.  If 
zinc,  for  instance,  were  placed  in  a  closed  electric  furnace  filled 
with  indifferent  gas  and  capable  of  supporting  100  atmospheres 
pressure,  zinc  could  be  kept  in  the  liquid  state  up  to  some 
1700°  C.,  and  its  alloying  with  other  metals  greatly  facilitated. 
In  the  making  of  brass,  for  instance,  much  zinc  is  lost  by  vola- 
tilization at  the  melting  point  of  the  copper  used,  yet  a  pressure 
on  the  furnace  of  four  atmospheres  would  entirely  prevent 
loss  of  zinc  at  the  melting  point  of  copper.  Induction  electric 
furnaces  could  easily  be  run  inside  a  pressure  vessel  at  this 
temperature,  and  the  pressure  be  relieved  gradually  as  the 
temperature  of  the  melted  alloy  was  reduced.  To  keep  zinc 
from  boiling,  or  to  keep  it  in  the  liquid  state,  at  1300°,  not  a 
high  temperature,  would  require  a  pressure  of  nearly  14  at- 
mospheres to  prevent  it  from  boiling  ad  libitum. 

Problem  135. 

Roasted  zinc  sulphide  ore,  consisting  practically  of  pure 
ZnO,  is  reduced  by  carbon,  and  the  reduction  gases  passed  into 
a  condenser.  Assume  the  reaction  to  be 

ZnO  +  C  =  Zn  +  CO. 

Required : 

(1)  The  temperature  at  which  zinc  commences  to  condense 
from  the  above  gas  mixture. 

(2)  The  proportion  of  the  zinc  condensed  out  for  each   1° 
reduction  of  temperature  below  this  "  dew  point." 

(3)  The  proportion  of  the  zinc  escaping  uncondensed,  if  the 
gas  escapes  from  the  condenser  at  600°  C. 

(4)  How  would  these  data  be  affected,  if  the  reduction  was 
taking  place  at  Denver,  1500  meters  above  sea  level,  barometer 
560  mm? 

Solution : 

(1)  Since  zinc  vapor  has  been  shown  to  have  a  specific  grav- 
ity of  32.5  referred  to  hydrogen  gas  at  the  same  temperature 
and  pressure,  and  the  molecular  weight  of  hydrogen  gas,  for- 
mula H2,  is  2,  the  molecular  weight  of  zinc  vapor  must  be 
2X32.5  =  65.  This  coincides  with  the  atomic  weight  of  zinc, 
and  therefore  zinc  vapor  is  monatomic  and  the  symbol  of  its 
molecule  is  Zn.  The  gas  mixture  in  the  above  equation  con- 


648  METALLURGICAL  CALCULATIONS. 

tains,  therefore,  one  molecule  each  of  its  constituents,  and, 
therefore,  each  gas  supports  half  the  atmospheric  pressure. 
This  is  ordinarily  expressed  by  saying  that  the  gas  is  half  one 
constituent  and  half  the  other,  meaning  that  if  the  two  gases 
were  separated  and  each  measured  under  the  prevailing  normal 
pressure,  the  volume  of  each  would  be  half  the  volume  of  the 
gas  mixture.  The  statement  that  each  supports  half  the  nor- 
mal pressure  is  the  more  scientific  and  logical,  for  in  a  gas 
mixture  each  gas  certainly  possesses  the  volume  of  the  mixture, 
but  under  only  a  fraction  of  the  pressure  on  the  mixture,  said 
fraction  being  identical  with  the  fraction  of  the  volume  of  the 
whole  which  it  would  constitute,  if  measured  separately  under 
the  pressure  supported  by  the  mixture. 

Consulting  the  table,  we  find  zinc  vapor  to  have  a  pressure  of 
380  mm.  at  862°.  This  would,  therefore,  be  the  dew  point,  at 
which  the  zinc  would  commence  to  condense.  (1) 

(2)  At  this  condensing  temperature,  a  difference  of  19°  in 
the  temperature  corresponds  to  82  mm.  difference  in  the  maxi- 
mum vapor  tension,  and,  therefore,  in  this  gas  mixture  at  this 
temperature,  a  reduction  of  1°  in  its  temperature  will  reduce 
the  tension  of  the  zinc  vapor  82-7-19  =  4.3,  or  practically  4  mm. 

4 

This  will  result  in  the  condensation  of  -^—^  th.  of  the  zinc,  be- 

ooU 

cause  its  tension  was  380  and  was  reduced  to  376,  and,  there- 

376 
fore,  -r^r-  th.  of  it  remained  uncondensed.     The  proportion  con- 

OoU 

densed  for  the  reduction  of  1°  in  temperature  is,  therefore,  a 
trifle  over  1  per  cent.  (2) 

(3)  Leaving  the  condenser  at  600°,  the  tension  of  the  zinc 
vapor  escaping  would  be,  from  the  table,  11.6  mm.     The  ten- 
sion of  the  CO  gas  would,  therefore,  be  760-  11.6  =  748.4  mm. 
For  every  cubic  meter  of  mixed  gases,  at  862°,  containing  a 
cubic  meter  of  zinc  vapor  at  that  temperature  and  380  mm. 
pressure,  there  will  escape  a  fraction  of  a  cubic  meter  of  mixed 
gas  at  600°,  containing  zinc  vapor  at  11.6  mm.  pressure.     The 
fractional  volume  can  be  calculated  from  the  tension  on  the  CO. 

1  m3  CO  at  862°  and  380  mm.  pressure,  reduced  to  600°  and 
748.4  mm.  pressure,  becomes 


THE  METALLURGY  OF  ZINC,  649 

600  +  273       380     _ 
X862  +  273X74O  ~ 

This  is,  then,  the  actual  volume  of  gas  mixture  escaping,  for 
each  actual  1  m3  of  gas  mixture  in  the  condenser  at  the  "  dew 
point,"  862°. 

The  uncondensed  zinc  vapor  is,  therefore,  0.381  m3  at  600° 
and  11.6  mm.  tension.  This  would  be,  at  862°  and  380  mm. 
tension 

862  +  273     11.6  _ 
°'3i  LX600  +  273XW 

There,  therefore,  escaped  uncondensed 

0.015  =  1.5  per  cent  of  the  zinc.  (3) 


A  quicker  solution,  not  so  easy  to  understand,  however,  is 
11.6 


748.4 


=  -0.015  =  1.5  per  cent  uncondensed. 


(4)  If  the  operation  takes  place  at  such  elevation  above  sea- 
level  that  the  barometer  stands  normally  at  560  mm.,  then  the 
partial  pressure  of  the  zinc  vapor  in  above  case  would  be  280 
mm.  instead  of  380  mm.  and  the  "  dew  point  "  or  temperature 
at  which  condensation  commences  would  be  835°  instead  of  862°, 
as  under  previous  conditions. 

At  this  temperature,  a  difference  of  1°  in  the  temperature 
corresponds  to  a  difference  of  3  mm.  in  the  tension  of  the  zinc 
vapor,  producing  therefore  a  condensation  of  the  zinc  present, 
per  1°  fall  of  temperature,  of  3-^-280  =  0.011  =  1.1  per  cent, 
instead  of  1.2  per  cent. 

If  the  temperature  of  the  escaping  gases  is  600°,  with  the 
saturated  zinc  vapor  at  11.8  mm.  tension  and  the  carbonous 
oxide,  CO,  at  560—11.8  =  548.2  mm.,  the  proportion  of  the 
original  quantity  of  zinc  escaping  uncondensed  is  11. 8-5-  548.2  = 
0.022  =  2.2  per  cent,  instead  of  1.5  per  cent. 

Condensation  of  Zinc  and  Mercury  Vapors. 

The  temperature  at  which  a  metallic  vapor,  like  zinc  or  mer- 
cury, commences  to  condense,  depends  upon  the  vapor  tension 
curve  of  the  metal  and  the  amount  of  other  uncondensable  gas 


650  METALLURGICAL  CALCULATIONS. 

with  Which  it  is  mixed.  These  intermixed  gases  reduce  the 
pressure  upon  the  metallic  vapor,  and  lower  the  temperature  at 
which  the  vapor  becomes  saturated  vapor.  The  phenomenon  is 
exactly  analogous  to  the  "  dew  point  "  of  air  and  the  precipita- 
tion of  rain. 

Problem  136. 

A  roasted  zinc  ore  contains  15  per  cent.  Fe2  O3  and  70  per  cent. 
Zn  O.     It  is  reduced  by  excess  of  carbon,  in  a  retort. 

Required: 

(1)  The  average  composition,  by  volume,  of  the  gas  mixture 
entering  the  condenser,  assuming  no  CO2  in  it. 

(2)  The  "  dew  point  "  of  the  mixture  at  which  it  begins  to 
deposit  zinc. 

(3)  The  percentage  of  zinc  escaping  condensation,  if  the  gases 
leave  the  condenser  at  600°  C. 

Solution: 

(1)  Per  kilogram  of  zinc  there  is  used,  assuming  complete 
reduction: 

1-7- (0.70X65/81)  =  1.78kg. 

Oxygen  in  1.78  kg.  of  ore: 

As  Zn  O  =  1X16/65  =  0.246  kg. 
As  Fe2O3  =  1.78X0.15X48/160  =  0.080  " 

Sum  =  0.326   " 

Co  formed  =  0.326X28/16  =  0.57  " 
Volume  of  CO  =  0.57-^-1.26  =  0.45m3 
Volume  of  Zn  =  1.00-7-2.93  =  0.34  m3 


Sum  =  0.79  m3 
Percentage  composition  of  gases : 

CO  =  57  per  cent. 
Zn  =  43       *          (1) 

(2)  If  the  barometric  pressure  is  assumed  normal,  then  the 
zinc  vapor  supports 

760X0.43  =  327mm 

and  the  temperature  at  which  the  mixture  becomes  saturated 
with  zinc  vapor  is  found  from  the  table  on  page  620  to  be  847°. 
This  is  83°  below  the  normal  boiling  point  of  zinc. 


THE  METALLURGY  OF  ZINC.  651 

(3)  From  847°  down,  the  vapor  in  the  retort  will  be  always 
saturated,  its  tension  being  found  from  the  tables.  The  propor- 
tion of  original  zinc  condensed,  or  remaining  uncondensed, 
cannot  be  inferred  simply  from  the  vapor  tension  at  any  tem- 
perature. For  instance,  the  vapor  tension  of  the  zinc  at  600°  C. 
is  12  mm  of  mercury  column.  The  CO  gas  leaving  the  con- 
denser at  this  temperature  will,  therefore,  be  at  a  tension  of 
760—12  =  748  mm,  and  will  be  accompanied  by  12/748  of  its 
volume  of  zinc  vapor;  as  it  came  into  the  condenser  it  was  ac- 
companied by  327/433  of  its  volume  of  zinc  vapor.  The  ratio 
of  these  two  quantities  or  proportions  represents  the  real  frac- 
tion of  the  zinc  escaping  uncondensed;  that  is,  the  proportion 
escaping  condensation,  in  terms  of  the  total  zinc  concerned,  is 

12/748-^327/423  =  0.016 -f- 0.773  =  0.008  =  0.8  per  cent. 

The  same  result  can  be  calculated  in  several  ways.  One  is 
based  on  the  datum  that  at  any  given  temperature  zinc  vapor 
is  65-^28  =  2.3  times  as  heavy  as  CO  gas. 

Problem  137. 

Numerous  attempts  have  been  made  to  reduce  Zn  O  continu- 
ously in  a  blast  furnace,  and  condense  the  zinc  from  the  gases. 
In  such  a  case,  the  gases  will  consist  of  zinc  vapor,  carbon 
monoxide  and  nitrogen,  and  the  deficit  of  reduction  heat  must  be 
obtained  by  the  burning  of  carbon  to  CO  at  the  region  of  the 
tuyeres.  A  low  charge  column  must  be  used  in  order  that  the 
gases  pass  out  hot  enough  to  carry  the  zinc.  Assume  the  gases 
to  escape  at  900°  C.,  and  the  furnace  to  lose  by  radiation  and 
conduction,  etc.,  an  amount  of  heat  equal  to  the  sensible  heat  in 
the  blast  used. 

Required: 

(1)  The  amount  of  fixed  carbon  to  be  charged  into  the  fur- 
nace, per  2000  Ib.  of  zinc. 

(2)  The  composition  of  the  gases  leaving  the  furnace. 

(3)  The  temperature  at  which  these  gases  will  begin  to  de- 
posit zinc  or  become  saturated  with  zinc  vapor. 

(4)  The  proportion  of  the  zinc  they  carry  which  will  escape 
from  the  condensers,  as  vapor,  at  600°  C. 

(5)  The  pressure  necessary  to  apply  to  the  furnace  to  keep 


652  METALLURGICAL  CALCULATIONS. 

the  zinc  in  the  melted  state  at  the  minimum  temperature  of  re- 
duction, 1033°  C. 

Solution: 

(1)  If  the  reaction  is  assumed  as  practically 

ZnO  +  C  =  Zn  +  CO 

there  is  needed  for  reduction,  per  2000  Ib.  of  zinc 
2000  X  (12/65)  =  369  Ib. 

There  is  in  this  reaction  a  deficit  of  heat,  which  is  per  65  parts 
of  zinc  concerned  simply 

(Zn,  O)-(C,  O)  =  84,800-29,160  =  55,640. 

This  deficit  can  only  be  made  up  by  burning  more  C  to  CO 
right  at  the  tuyeres.  If  two  more  atoms  of  carbon  were  burned 
we  would  have  a  little  more  than  enough  heat.  We  will,  there- 
fore, assume  the  reaction  at  the  tuyere  region  as  the  using  of 
one  atom  of  carbon  for  reduction  and  of  double  as  much  for 
supplying  the  heat  deficit,  and  since  one  volume  of  oxygen  is 
accompanied  by  3.81  volumes  of  nitrogen,  the  whole  reaction 
will  be 

Zn  O  +  3  C  +  02  +  3.81  N2  =  Zn  +  3  CO  +  3.81  N2. 

The  amount  of  fixed  carbon  necessary  is  approximately  three 
times  that  necessary  for  reduction  only  and  amounts,  per 
2000  Ib.  of  zinc,  to 

36X~22=11081b.  (1) 

(2)  The  furnace  gases,  therefore,  contain,  by  volume, 

Zinc  vapor  =  1  volume  =  12.8  per  cent. 

Carbon  monoxide      =3  "       =  38.4       " 

Nitrogen  =3.81  "      =  48.8       "  (2) 

(3)  The  gases  will  be  saturated  with  zinc  vapor  when  the 
partial  pressure  of  the  zinc,  which  is 

760X0.128  =  97mm 

is  equal  to  the  maximum  tension  of  zinc  at  the  temperature 
attained.  This  is  at  745°  C.,  as  seen  from  inspecting  the  vapor 
tension  tables,  or  185°  below  the  normal  boiling  point  of  zinc.  (3) 


THE  METALLURGY  OF  ZINC.  653 

(4)   The  relative  volume  of  zinc  vapor  to  uncondensable  gas 
in  the  original  gases  is 


In  the  gases  issuing  from  the  condenser  at  600°  it  would  be 

w  -  «•«• 

The  proportion  of  the  zinc  escaping  condensation  under  these 
conditions  would,  therefore,  be 

!L°16  =  0  .  109  =  10  .  9  per  cent.  (4) 

By  cooling  the  gases  to  the  melting  point  of  zinc,  it  would 
be  possible  to  leave  uncondensed  only 

=  0.0014  =  0.14  per  cent. 

(5)  If  it  was  a  question  only  of  keeping  pure  zinc  from  vaporiz- 
ing at  1033°,  in  the  absence  of  other  gases,  we  could  at  once 
consult  the  vapor  tension  table,  and  find  2.2  atmospheres 
as  the  actual  tension  of  zinc  vapor  at  1033°,  corresponding  to 
1.2  atmospheres  effective  pressure  upon  the  furnace.  But  in 
the  gas  mixture  coming  from  the  furnace  the  zinc  is  practically 
one-eighth,  by  volume,  of  the  gases,  which  means  that  it  sus- 
tains one-eighth  of  '  the  pressure  upon  the  whole.  It  would, 
therefore,  take  a  tension  of 

2.2X8  =  17.6  atmospheres 

to  keep  all  this  zinc  in  the  liquid  state,  corresponding  to  an 
effective  pressure  of  16.6  atmospheres.  (5) 

This  is  an  entirely  impracticable  working  condition  and 
shows  the  practical  impossibility  of  the  schemes  upon  which 
much  money  have  been  spent  for  reducing  zinc  ore  to  liquid 
zinc  in  a  blast  furnace  run  under  high  pressure. 

Problem  138. 

In  the  metallurgy  of  mercury,  the  average  ore  contains  2 
per  cent,  of  Hg  S  and  is  roasted  by  the  use  of  10  per  cent,  of  its 


654  METALLURGICAL  CALCULATIONS. 

weight  of  wood  having  an  average  composition  of  water  20 
per  cent.,  carbon  32,  hydrogen  5.3,  oxygen  42.7.  Assume  that 
just  enough  air  enters  to  completely  burn  the  sulphur  and  car- 
bon, that  the  ore  and  air  used  are  dry. 

Required: 

(1)  The  percentage  composition  by  volume  of  the  roaster 
gases. 

(2)  The  temperature  at  which  the  mercury  will  commence 
to  condense. 

(3)  The    proportion    of    the    mercury    present    escaping    as 
vapor  if  the  gases  pass  out  of  the  condensers  at  100°  C.,  at 
50°  C.,  at  15°  C. 

Solution: 

(1)  Per  1000  kg.  of  ore,  containing  20  kg.  of  Hg  S,  there 
must  be  burned 

20X32/232  =    2.8  kg  of  sulphur. 
100X0.32       =  32       "    "    carbon. 

requiring  of  oxygen : 

2.8X1  =      2.8kg. 

32     X 32/12  =    85.3    " 


Sum  =    88.1 
Nitrogen  accompanying  =  293 . 7 


Air  =  381.8  ' 

Composition  of  the  gases: 

Hg    =    17.2  kg  =    17.2  ^  9.00  =      1.91  m3  =  0.5    per  cent. 

SO2  =      5.6   "   =      5.6-7-2.88=      1.91    "   =  0.5 

H2O=    68.0   "   =    68.0 -=-0.81=    83.95    "   =  22.1 

N2    =  293.7    "   =  293.7  -v- 1.26  =  233.10   "   =  61.3 

CO2  =  117.3    "   =  117.3  ^  1.98  =    59.24   "   =  15.6 


380.11    "   =  100.00  (1) 

(2)  In  reality,  each  of  these  gases  possesses  the  volume  of  the 
gas  mixture,  and  is  at  the  above  percentage  of  the  normal 
barometric  pressure  upon  the  mixture.  The  tensions  of  the 
different  constituents  of  the  mixture  are,  therefore, 


THE  METALLURGY  OF  ZINC.  655 

Hg    760X0.005  =      3.8mm 
SO2   760  X  0.005  =      3.8     " 
H2O  760  X  0.221  =  168.0     " 
N2     760  X  0.613  =  465.9    " 
CO2  760X0.156  =  118.5     " 

760.      " 

The  mercury  will,  therefore,  commence  to  condense  when  the 
temperature  of  the  gas  mixture  is  that  of  mercury  vapor  at  a 
maximum  tension  of  3.80  mm.  From  the  vapor  tension  table 
of  mercury  we  find  this  to  be  156°  C.,  or  201°  below  the  nor- 
mal boiling  point  of  mercury.  Above  this  temperature  the  gas 
mixture  is  not  saturated  with  Hg  vapor  and  no  condensation 
can  occur.  (2) 

(3)  In  the  gas  mixture,  the  mercury  vapor  is  equivalent  to 
3.8/760  of  the  volume  of  the  whole,  or  3.8/756.2  of  the  volume 
of  the  other  gases.  This  latter  proportion  is  0.00503.  At 
100°  C.,  the  uncondensed  mercury  vapor  can  have  a  tension  of 
0.285  mm,  leaving  759.715  to  the  other  gases,  and  giving  the 
ratio  of  mercury  vapor  to  other  gases  0.000375.  Since  the  other 
gases  have  remained  constant  in  amount,  the  proportion  of 
mercury  remaining  condensed  is 

0.000375  -f-  0.00503  =  0.075  =  7.5  per  cent. 

If  the  temperature  of  the  mixture  is  reduced  to  50°  C.,  the 
mercury  vapor  remaining  can  exert  a  tension  of  only  0.013  mm 
and  the  water  vapor  only  92  mm.  The  other  gases  present  will 
therefore  sustain  760-  (92  +  0.013)  =  668  mm.  In  the  origi- 
nal mixture  the  Hg  and  H2O  vapors  are,  respectively,  equiva- 
lent to 

Hg        3.8  ^  587.2  =  0.0065 
H2O  168.0  -^  587.2  =  0.2860 

of  the  volume  of  N2,  CO2  and  SO2  together. 

In  the  gas  escaping  at  50°  the  Hg  and  H2O  will  be  equiva- 
lent to 

Hg      0.013^-668  =  0.00002 
H2O92.        -668  =  0.1377 

of  the  volume  of  the  same  gases. 

The  proportions  of  the  Hg  and  H2O  escaping  uncondensed 
at  50°  will,  therefore,  be 


656  METALLURGICAL  CALCULATIONS. 

Hg     0.00002  -J-  0.0065  =  0.0031  =    0.31  per  cent. 
H2O  0.1377    -f-  0.2860  =  0.4815  =  48.15  per  cent. 

By  exactly  similar  reasoning,  using  the  maximum  tensions  of 
Hg  and  H2O  at  15°  (0.0009  mm  and  13  mm,  respectively),  the 
proportions  of  each  to  the  N2  +  CO2  +  SO2  are 

Hg      0.0009  -f-  747  =  0.0000012 
H2O  13.          -f-  747  =  0.0174 

>;•'•*  "  •      '        .  1" 

and  the  proportions  escaping  condensation  would  be  ; 

Hg    0.0000012  -f-  0.0065  =  0.00018  =  0.018  per  cent. 
H2O  0.0174       -f-  0.2860  =  0.06064  =  6.064  per  cent.      (4) 

The  differences  between  these  percentages  and  100  .will  be 
the  proportions  of  these  materials  condensed  to  the  liquid  state. 

METALLIC  MIST  OR  FUME. 

When  zinc  or  mercury  are  in  the  state  of  vapor  their  atoms 
are  each  separate  from  other  atoms.  Chemically  speaking, 
their  vapor  molecules  are  mon-atomic.  On  condensing  to  the 
liquid  state  we  do  not  know  whether  the  atoms  come  together 
into  compound  molecules  or  whether  they  simply  come  closer 
together.  In  either  case,  the  molecules  of  the  liquid  metal,  as 
the  temperature  is  reduced  low  enough  to  form  them,  exist  at 
first  as  isolated  molecules,  constituting  the  liquid  metal  in  its 
finest  possible  state  of  sub-division,  so  small  that  it  is  for  the 
time  being  practically .  still  a  gas.  If,  however,  we  give  these 
liquid  molecules  the  possibility  of  uniting  to  liquid  masses,  we 
get  the  latter.  This  is  a  matter  principally  of  reducing  the  sur- 
face tension  which  holds  the  liquid  molecules  each  to  itself  as  a 
sphere.  Contact  with  a  rubbing  surface,  with  dust  particles, 
filtration  through  cloth,  or  even  electrification  (perhaps  a  mag- 
netic field)  reduce  this  surface  tension  and  cause  agglomera- 
tion into  liquid  masses  which  precipitate. 

The  amount  of  liquid  or  solid  particles  thus  held  in  suspen- 
sion and  escaping  with  the  current  of  uncondensable  gas  is  en- 
tirely independent  of  the  quantity  escaping  as  true  vapor,  which 
has  been  calculated  above,  except  in  so  far  as  they  are  both 
functions  of  the  amount  of  said  uncondensable  gas.  The  veloc- 
ity of  the  escaping  current  is  a  large  factor  in  the  amount  of 
mist,  because  the  carrying  power  of  gas  for  mist  particles  varies 


THE  METALLURGY  OF  ZINC.  657 

probably  with  the  cube  of  its  velocity,  so  that  halving  the 
velocity  of  the  issuing  gas  will  diminish  greatly  this  source 
of  loss.  Large  settling  chambers  in  which  the  mist  can  de- 
posit, because  of  stagnation  of  the  gas  current,  and  large  rub- 
bing surfaces,  are  effective  means  for  catching  or  depositing 
the  mist.  Filtration  through  bags  is  still  more  effective. 

In  the  case  of  zinc,  the  solidifying  point  is  420°,  while  con- 
densation from  the  state  of  vapor  begins  at,  say,  860°.  In  this 
interval  only  can  the  condensed  particles,  as  a  mist,  collect 
together  to  liquid  zinc.  If  the  temperature  passes  quickly 
through  this  range,  the  particles  have  little  chance  to  agglomer- 
ate into  liquid  masses,  and  the  proportion  which  chills  into 
solidified  mist  particles  is  increased.  These  solidified  mist 
particles,  analogous  to  "  hoar  frost  "  in  nature,  form  the  well 
known  "  blue  powder  "  of  the  zinc  works.  It  settles  in  large 
settling  chambers,  and  is  in  extremely  fine  particles,  mostly 
under  0.01  mm  in  size.  It  is  quite  easy  to  collect  all  the  zinc 
in  this  form  if  the  vapor  is  chilled  suddenly  and  the  resulting 
gas  settled  properly  or  filtered. 

The  loss  of  mercury  or  zinc  as  mist  or  hoar  frost  is  there- 
fore, to  be  considered  entirely  apart  from  the  loss  as  true  vapor; 
it  may  be  smaller  than  the  latter  and  it  may  be  larger ;  its  amount 
varies  particularly  with  the  nature  of  the  settling  apparatus 
and  the  velocity  of  the  gases  as  they  escape.  The  theory  as 
to  its  amount  under  any  given  set  of  conditions  would  be  very 
difficult  to  elaborate,  and  the  data  for  doing  so  with  exactness 
are  practically  unknown.  The  best  that  can  be  done  at  present 
is  to  study  the  loss  as  mist  or  hoar  frost  in  actual  practice, 
experimentally,  and  tabulate  the  actual  results  as  a  guide  for 
future  metallurgical  use. 


CHAPTER  V. 
METALLURGY  OF  ALUMINIUM. 

The  two  essential  principles  here  involved  are  "  differential 
reduction  "  as  used  in  the  electric  furnace  purification  of  alumina, 
and  "  electrolytic  furnace  operation,"  as  illustrated  in  the 
decomposition  of  the  alumina  by  electrolysis  in  the  manner 
usually  practiced.  Only  the  latter  problem  will  be  covered  here. 

ELECTROLYTIC  FURNACE  REDUCTION  OF  ALUMINA. 

The  Hall  process  is  beautifully  simple  and  technically  admir- 
able. Al2  O3  is  found  to  dissolve  in  melted  alkaline-aluminium 
double  fluorides;  it  is  as  pretty  a  case  of  solution,  so  far  as  ap- 
pearances go,  as  dissolving  a  spoonful  of  powdered  sugar  in  a 
glass  of  distilled  water.  The  melting  point  of  the  fused  fluorides 
is  reduced  by  the  solution  of  alumina,  just  as  that  of  water  is 
reduced  by  dissolving  salt.  In  passing  the  electric  current  the 
constituents  of  the  dissolved  alumina  appear  at  the  electrodes, 
oxygen  at  the  anode  and  aluminium  at  the  cathode.  The  best 
practical  arrangement  is  to  use  a  carbon-lined  pot,  with  molten 
aluminium  in  the  bottom  as  the  cathode,  upon  it  a  few  inches 
depth  of  the  bath,  and  dipping  into  this  the  carbon  anodes.  The 
writer  has  given  most  of  the  technical  details  of  this  operation 
in  his  treatise  on  "Aluminium,"  and  Prof.  Haber  has  published 
extensive  laboratory  studies  of  the  process  in  the  Zeitsckrift 
fur  Elektrochemie. 

With  a  solvent  salt  consisting  of  melted  sodium  fluoride  and 
aluminium  fluoride,  such  as  called  for  in  one  of  the  Hall  patents, 
with  alumina  dissolved  therein,  and  using  carbon  anodes,  the 
electrolytic  elements  of  the  process  are  simplicity  itself.  The 
bath  contains  sodium,  aluminium,  fluorine  and  oxygen,  and  the 
anode  is  carbon.  Under  these  conditions  those  elements  or  com- 
pounds will  form  at  the  electrodes  which  cannot  further  react 
upon  the  bath  material;  in  other  words,  those  materials  most 
stable  in  contact  with  the  bath  material  or  electrodes.  A  mo- 

658 


THE  METALLURGY  OF  ALUMINIUM.  659 

ment's  reflection  explains  what  happens,  and  what  must  happen. 
At  the  cathode,  sodium  cannot  be  liberated  because  metallic 
sodium  reacts  chemically  on  this  bath,  separating  out  alumin- 
ium; therefore,-  the  electrolytic  reducing  tendency  at  the  sur- 
face of  the  cathode  can  only  expend  itself  in  separating  out 
aluminium.  At  the  anode,  fluorine  cannot  be  liberated  because 
fluorine  acts  strongly  upon  alumina  even  when  cold,  converting 
it  into  fluoride  and  expelling  its  oxygen;  therefore,  the  electro- 
lytic perducing  tendency  at  the  surface  of  the  anode  will  tend 
to  set  free  oxygen.  But  oxygen  cannot  be  set  free  at  a  carbon 
surface  at  a  cherry-red  heat  because  of  its  inevitable  tendency 
to  unite  with  the  carbon  to  form  CO.  The  electrolytic  agency 
at  the  surface  of  the  carbon  anode  must,  therefore,  cause  the 
formation  of  CO.  The  whole  electrolytic  operation  results  in 
the  removal  of  A12O3  from  the  bath  and  the  formation  of  alumin- 
ium and  carbon  monoxide. 

The  thermochemical  relations,  as  far  as  known,  agree  abso- 
lutely with  the,  above  explained  experimental  results.  The 
materials  in  presence  of  each  other  are  sodium  fluoride,  alumin- 
ium fluoride,  aluminium  oxide  and  carbon.  The  heats  of 
formation  of  these,  per  molecule  and  per  chemical  equivalent 
concerned,  are  as  follows: 

(Na,  F)  =  109,720  '=  109,720  per  chemical  equivalent. 

(Al,  F3)  =  275,220  =     91,740    " 

(Al2,  O3)  =  392,600  =     65,430    " 

(C,  O)  =    29,160  =     14,580    " 

The  heat  of  formation  of  carbon  tetra-fluoride  is  unknown, 
but  is  probably  small,  since  it  is  so  difficult  to  form. 

From  the  last  column,  which  largely  governs  the  work  done 
by  the  current,  we  see  that  the  current  does  far  the  least  work 
when  it  decomposes  alumina;  in  fact,  it  does  still  less,  by  14,580 
calories,  because  of  the  assistance  rendered  by  carbon  uniting 
with  the  oxygen.  Now,  although  the  electrical  current  does  not 
consistently  adhere  to  the  doctrine  of  "  least  work,"  yet  it  does 
in  this  case  because  forced  to  do  so  by  the  chemical  relations 
of  sodium,  aluminium,  fluorine,  oxygen  and  carbon  at  the  tem- 
perature of  the  bath,  as  explained  in  the  preceding  analysis. 
It  is  probable  that  in  electrolysis  the  chemical  relations  of  the 
possible  products  control  what  the  current  does  rather  than  the 


660  METALLURGICAL  CALCULATIONS. 

thermochemical  relations  alone,  but  in  most  cases  the  two  condi- 
tions or  controlling  circumstances  coincide  in  their  influence 
and  lead  to  identical  results.  This  is  the  case  in  the  process 
in  question ;  it  is  absolutely  normal  and  is  explainable  by  either 
method  of  reasoning. 

If  the  operation  is  run  with  a  small  vessel  and  correspond- 
ingly small  current,  the  heat  necessarily  evolved  by  the  passage 
of  the  current  is  small  compared  to  the  conduction  and  radia- 
tion losses  and  must  be  supplemented  by  outside  heating  to 
keep  the  contents  at  proper  temperature.  If  the  size  of  the 
operation  is  increased  in  all  its  items  and  dimensions,  the  neces- 
sarily generated  internal  "  resistance  "  heat  will  suffice  to  keep 
the  bath  at  the  requisite  temperature,  when  the  enlargement  is 
done  on  a  certain  scale.  If  enlarged  past  this  point,  too  much 
internal  heat  is  unavoidably  generated  and  means  must  be  used 
to  artificially  cool  the  pot. 

Problem  139. 

An  electrolytic  vessel  is  composed  of  a  block  of  carbon  25  cm 
cube,  with  a  cavity  10  cm  square  by  10  cm  deep  inside.  The 
cavity  has  a  round  carbon,  5  cm  diameter,  dipping  into  it.  The 
vessel  weighs  30  kilograms;  the  fused  bath  in  the  cavity  2  kg; 
the  carbon  immersed  in  it  0.1  kg.  The  specific  heat  of  the  car- 
bon is  0.5,  of  the  bath  0.3,  at  the  running  temperature.  An  ex- 
periment showed  that  the  bath  material  cooled  off,  at  the  work- 
ing temperature,  at  the  rate  of  10°  C.  per  minute,  the  walls  of 
the  vessel  at  an  average  rate  of  2°  C.  per  minute,  when  left  to 
cool  by  themselves. 

Required: 

(1)  The  number  of  watts  which  must  be  converted  into  heat 
in  the  vessel  in  order  to  maintain  it  at  the  working  temperature. 

(2)  Assuming  75  per  cent,  of  the  theoretical  ampere  efficiency 
to  be  obtained,  what  amperes  passed  through  the  pot  will  keep 
it  at  working  temperature  if  the  working  voltage  is  kept  at  10 
volts? 

Solution: 

(1)  The  30  kg  of  vessel  material  losing  heat  at  the  rate  of 
2°  per  minute,  with  specific  heat  of  0.5,  gives  a  heat  loss  per 

minute  of 

30X0.5X2  =  30  Calories. 


THE  METALLURGY  OF  ALUMINIUM.  *  661 

Similarly,  the  immersed  carbon  in  the  cavity  and  the  bath 
material  itself  lose 

0.1X0.5X10  =  0.5  Calories. 
2.0X0.3X10  =  6.0 

The  total  heat  loss  is,  therefore,  36.5  Calories  per  minute. 

To  maintain  the  temperature  constant  the  current  must  fur- 
nish this  heat,  and  since  1  watt  is  0.239  gram  calories  per  sec- 
ond, the  watt  energy  thus  converted  into  heat  must  be 

36.5X1000       0-,- 
0.239X60   ==2545watts  (D 

(2)  If  all  the  amperes  passing  through  separated  out  metal 
the  voltage  absorbed  in  decomposition  in  the  bath  would  be 
from  the  thermochemical  heats  of  formation  of  chemical  equiva- 
lent quantities  of  AP  O3  and  CO: 

65,430-14,580 

=  2.2  volts. 


23,040 

If  75  per  cent,  of  the  amperes  are  efficient,  the  voltage  thus  ab- 
sorbed is 

2.2X0.75  =  1.65  volts. 

The  voltage  disappearing  in  overcoming  ohmic  resistance  will 
then  be 

10-1.65  =  8.35  volts 

and  the  current  which  when  passed  will  keep  the  pot  at  working 
temperature  will  be 

2545 

g-^5  =  305  amperes.  (2) 

The  above-used  principles  are  applicable  to  any  kind  of  elec- 
trolytic-furnace operation. 


INDEX. 

ADDICKS,  L.,  on  energy  loss  at  electrical  contacts 562 

Air,  composition  of 4 

Air,  received  by  converter 336 

Air,  required  by  converter 333 

Alloys,  heats  of  formation 37 

Alloys,  thermophysics  of 109 

Alumina,  action  in  blast  furnace  slag 256 

Alumina,  behavior  in  the  blast  furnace 240 

Alumina,  reduction  of 658 

Alumina,  thermophysics  of 123 

Aluminates,  heats  of  formation  of 35 

Aluminium  alloys,  thermophysics  of 113 

Aluminium  compounds,  heats  of  formation  of 18,  37 

Aluminium  compounds,  thermophysics  of 139 

Aluminium  conductors,  economical  size  of 570 

Aluminium  conductors,  heating  of,  in  use ' 572 

Aluminium,  electrical  conductivity  of 569 

Aluminium,  heat  of  oxidation  of 353,  374 

Aluminium,  metallurgy  of 658 

Aluminium,  reduction  of  lead  oxide  by 586 

Aluminium,  thermophysics  of 80 

Amalgam  furnace,  efficiency  of , 107 

Amalgams,  heats  of  formation  of 37 

Ammonia  gas,  mean  specific  heat  of 142 

Anode,  insoluble,  use  of 551 

Anode,  soluble,  use  of  iron  as 548 

Anode,  use  of  matte  as 539 

Anthracite  coal,  combustion  of 220,  221 

Antimonides,  heats  of  formation  of 23 

Antimonides,  thermophysics  of , 136 

Antimony  alloys,  thermophysics  of 101 

Antimony,  behavior  in  the  blast  furnace 240 

Antimony,  compounds,  heats  of  formation  of 23 

Antimony,  thermophysics  of 95 

Aqueous  vapor,  maximum  tension  of ' Ill 

Argo,  Col.,  reverberatory  smelting  at 501 

Arsenic,  behavior  in  the  blast  furnace 240 

Arsenic  compounds,  heats  of  formation  of 33 

Arsenic  compounds,  thermophysics  of 136, 138 

Arsenic,  thermophysics  of 88 

Arsenides,  heats  of  formation  of 23 

Ashcroft's  process  for  copper  ores 537 

Atacamite,  formula  of 537 

Atomic  weights 1 

BALANCE  sheet  of  blast  furnace 235 

Barium  compounds,  heats  of  formation  of 18-40 

Barium  compounds,  thermophysics  of 113-144 

Barium,  thermophysics  of 97 

Bauxite,  efficiency  of  furnace,  in  melting 127 

Beeswax,  thermophysics  of 142 

Belgian  furnace,  distillation  of  zinc  ore  in  a 635 

Bergfeld,  on  the  boiling  of  silver  and  gold 615 

663 


664  INDEX 

Beryllium  compounds,  heats  of  formation  of 18-40 

Beryllium  compounds,  thermophysics  of 113-144 

Beryllium,  thermophysics  of 76 

Bessemerizing  copper  matte 525 

Bessemer  process,  problem  concerning 11 

Bessemer  process,  the 333 

Betts'  refining  process  for  lead 598,  599,  602 

bi-carbonates,  heats  of  formation  of 29 

Bismuth  alloys,  thermophysics  of . . . . 110 

Bismuth  compounds,  heats  of  formation  of 18-40 

Bismuth  compounds,  thermophysics  of 113-144 

Bismuth,  thermophysics  of ; 102 

Bi-sulfates,  heats  of  formation  of • 33 

Bituminous  coal,  combustion  of 219,  221,  222,  223,  226 

Blast,  amount  used  in  blast  furnace 244 

Blast,  dry,  Gayley  process  of  producing , 330 

Blast,  dry,  the  rationale  of 309 

Blast  engines,  efficiency  of 493,  590 

Blast  furnace,  balance  sheet  of 235 

Blast  furnace,  carbon  needed  in  the •    277 

Blast  furnace  gas,  in  gas  engines 226 

Blast  furnace  gases,  surplus  power  from 272 

Blast  furnace,  heat  balance  sheet  of .  281 

Blast  furnace,  reduction  of  zinc  ores 651 

Blast  furnace,  temperature  in 223 

Blast  furnace,  utilization  of  fuel  in  the 268 

Blast,  heating  of,  apparatus  for 318 

Blast,  hot,  the  function  of 308 

Blast  pressure  required  in  Bessemer  process 340 

Blast,  warm,  use  in  pyritic  smelting 487 

Blast,  work  done  in  producing 313 

Blister  roasting  of  copper  matte 523 

Blowing  engines,  efficiency  of 299 

Boiling  of  metals  in  a  vacuum 614 

Boiling  of  zinc 641 

Borates,  heats  of  formation  of 34 

Boron  compounds,  heats  of  formation  of 18-40 

Boron  compounds,  thermophysics  of 138 

Boron,  thermophysics  of 77 

Brass,  thermophysics  of 

Bringing  forward  of  copper  matte .  514 

Bromides,  thermophysics  of 133 

Bromine,  thermophysics  of 89 

Bronze,  thermophysics  of Ill 

Bullion,  gold,  electrolytic  refining  of 606 

Bullion,  silver,  electrolytic  refining  of 605 

Burgess  process  of  refining  iron 

Butte,  Mont.,  copper  concentrates  from 479 

Butte,  Mont.,  roasting  furnace  at 482 

CADMIUM  alloys,  thermophysics  of 113 

Cadmium  compounds,  heats  of  formation  of 18-40 

Cadmium  compounds,  thermophysics  of 131-144 

Cadmium,  thermophysics  of 93 

Cadmium,  vapor  tension  of 644 

Caesium  compounds,  heats  of  formation  of 18-40 

Caesium,  thermophysics  of 96 

Calcium  compounds,  heats  of  formation  of 18-40 

Calcium  compounds,  thermophysics  of 131-144 

Calcium  oxide,  heat  absorbed  in  reducing 293 


INDEX  665 

Calcium  oxide,  thermpphysics  of . .         124 

Calcium,  thermophysics  of 83 

Calorific  power  of  fuels,  calculation  of 503 

Calorimeter,  test  for  specific  heat  by 225 

Calorimeter,  test  for  temperature  by 224 

Campbell,  H.  H.,  furnace  data  by 393 

Canada,  electrical  processes  in • 435 

Canada  Zinc  Co.,  operations  of  the 641 

Carbides,  heats  of  formation  of 25 

Carbon,  amount  needed  in  the  blast  furnace 277 

Carbon,  heat  of  oxidation  of 352,  375 

Carbon,  thermophysics  of 77 

Carbonates,  heat  absorbed  in  decomposing : .  : 291 

Carbonates,  heats  of  formation  of 28 

Carbonates,  thermophysics  of 137 

Carbonic  oxide,  thermophysics -of 122 

Carbonous  oxide,  thermophysics  of 121 

Cathodes,  use  of  copper  matte  as 545 

Cathodes,  use  of  lead  sulfide  as 598 

Cement  copper,  composition  of 548 

Cerium  compounds,  heats  of  formation  of 18,  40 

Cerium,  thermophysics  of 97 

Chalcopyrite,  composition  of ' 474 

Charge,  blast  furnace,  calculation  of . . . . : 250 

Chimney  draft .          185 

Chimney  draft,  problems  concerning 192, 194,  224,  230,  231 

Chimney  for  open-hearth  furnace 395 

Chlorates,  thermophysics  of . 139 

Chlorides,  heats  of  formation  of 27 

Chlorides,  thermophysics  of 131 

Chlorine,  thermophysics  of 82 

Chromates,  thermophysics  of 138 

Chromium,  heat  of  oxidation  of 353,  374 

Chromium,  thermophysics  of 84 

Coal,  combustion  of  powdered 

Cobalt,  behavior  in  the  blast  furnace 240 

Cobalt  compounds,  heats  of  formation  of 18-40 

Cobalt  compounds,  thermophysics  of 131-144 

Cobalt,  thermophysics  of 87 

Coehn,  improved  Hoepf ner  apparatus 558 

Coke-oven  gases,  combustion  of 219-228 

Coke,  use  of,  in  pyritic  smelting 488 

Columbium,  thermophysics  of • 90 

Compression  of  blast,  work  done  in 313 

Concentration  of  zinc  ores,  cost  of 621 

Condensation  of  metallic  vapors 649,  654 

Conduction  of  heat,  principles  of 200 

Conductivity  for  heat,  tables  of 203,  211 

Conductors,  electrical,  economic  size  of 564 

Conductors,  electrical,  heating  of,  in  use 568 

Constants,  thermochemical 38-40 

Contacts,  electric,  energy  loss  at • .          562 

Copper  alloys,  thermophysics  of Ill 

Copper,  behavior  in  the  blast  furnace 240 

Copper,  calculations  on  metallurgy  of 469,  619 

Copper  chloride,  electrolysis  of 537,  557 

Copper  compounds,  heats  of  formation  of 18-40 

Copper  compounds,  thermpphysics  of 131-144 

Copper,  electrical  conductivity  of ........: 567 

Copper,  electrolytic  refinery  of 558 


666  INDEX 

Copper  matte,  composition  of 469,  471 

Copper,  principles  of  the  metallurgy  of 472 

Copper  ores,  electric  smelting  of 572 

Copper  solutions,  physical  properties  of 611 

Copper  solutions,  resistivity  of :  .  .  .  559 

Copper,  thermophysics  of 87 

Crushing  of  zinc  ores,  cost  of 621 

Cupola,  thermal  efficiency  of : 115 

Cyanates,  heats  of  formation  of 36 

Cyanides,  heats  of  formation  of 35,  36 

Cyanides,  thermophysics  of 136 

DAMOUR,  on  furnace  efficiency , 416,  423 

Decomposition  of  moisture  of  blast 293 

Dehydrating  charges,  heat  absorbed  in 290 

Dell  wick-Fleischer  gas  producer 179 

Distillation  of  zinc  from  ore 632 

Dried  blast,  rationale  of  action  of 309 

Dried  blast,  temperatures  attained  with 312 

Drying  air  blast  commercially 325 

Drying  charges,  heat  absorbed  in 290 

Drying  of  wet  peat 227 

Ducktown  pyrrhotite,  smelting  of 489 

Dulong  and  Petit's  laws  of  specific  heats 70 

Dulong's  laws  of  heat  of  combustion  of  coals 48 

EFFICIENCY  of  electric  furnaces 640 

Efficiency  of  electric  heating 441 

Efficiency  of  furnaces,  discussion  of 104 

Efficiency  of  open-hearth  furnaces 396 

Eissler's  hydrometallurgy  of  copper 469 

Eldred  process  of  combustion 55 

Electric  furnaces,  efficiency  of 640 

Electric  melting  of  steel 440 

Electric  reduction  of  iron  ore 429 

Electric  refining  of  iron 446 

Electric  smelting  of  copper  ores 572 

Electric  smelting  of  zinc  ores 637 

Electrical  contacts,  energy  loss  at 562 

Electrolytic  furnace,  definition  of • 534 

Electrolytic  furnace  for  reducing  aluminium . .  .  .  • 658 

Electrolytic  refining  of  copper :  .  .  .  558 

Electrolytic  refining  of  gold 610 

Electrolytic  refining  of  iron 446 

Electrolytic  refining  of  lead 598 

Electrolytic  refining  of  silver 605 

Electrolytic  refining,  principles  of 536 

Electrometallurgy  of  copper 534 

Electrometallurgy  of  iron  and  steel • 429 

Electrometallurgy  of  lead '. 598 

Electrothermal  processes,  definition  of 535 

Ely,  Vermont,  copper  ore  from 475 

Ethylene,  mean  specific  heat  of 142 

Evans-Klepetko  furnace,  calculations  on 482 

FEED-WATER  heater,  utility  of 225 

Ferro-cyanides,  heats  of  formation  of 36 

Fire-clay,  heat  conductivity  of 630 

Flues,  size  of 382 

Fluorides,  heats  of  formation  of 26 


INDEX  667 

Fluorides,  thermophysics  of 134 

Flux,  amount  needed  in  the  blast  furnace 251 

Flux,  behavior  in  the  blast  furnace 243 

Flux,  comparison  of  values  of 263 

Flux,  required  in  Bessemer  converter 346 

Freeland,  W.  H.,  on  smelting  pyrrhotite.' 489,  515 

Fuels,  behavior  of,  in  the  blast  furnace 236 

Fuels,  calorific  power  of 46 

Fuels,  comparison  of  values  of 261 

Fuels,  utilization  of,  in  the  blast  furnace 268 

Fume,  metallic 656 

Furnaces,  efficiency  of 104 

Furnaces,  electric,  efficiency  of 640 

Furnaces,  electrolytic,  definition  of •  534 

Furnaces,  electrolytic,  for  reducing  alumina 658 

Fusion  of  latent  heats  of,  discussion 71 

Fusion  of  latent  heats  of,  tables 118-144 

Fusion  of  matte 500 

Fusion  of  slag 500 

GALENA,  roasting  of 581 

Gallium,  thermophysics  of 88 

Gas,  artificial 145 

Gas,  engine  operation 226 

Gas,  mixed,  producers  for 158 

Gas,  natural,  combustion  of 10,  47,  53,  219 

Gas,  natural,  from  lola,  Kansas 636 

Gas  producers 381 

Gases,  composition  of,  from  smelting  furnace 497 

Gases,  pressure  corrections  for 7 

Gases,  relative  volumes  of 

Gases,  temperature  corrections  for 5 

Gases,  thermophysics  of 141 

Gases,  waste,  heat  in 287 

Gayley  process  of  drying  blast 330 

Genoa,  Marchesi's  process  at 539 

Gin,  G.,  in  electrometallurgy  of  zinc 637 

Glass,  thermophysics  of 141 

Gold,  behavior  in  the  blast  furnace 240 

Gold  bullion,  refining  of 610 

Gold  compounds,  heats  of  formation  of 18—40 

Gold  compounds,  thermophysics  of 131-144 

Gold,  metallurgy  of 605 

Gold,  thermophysics  of 100 

Gold,  vapor  tension  of 616 

Gordon,  F.  W.,  slag  calculations  by 261 

Goutal's  method  for  calorific  power  of  coal 503 

Great  Falls,  Montana,  copper  refining  at 566 

Gruner's  ideal  working  of  a  blast  furnace 274 

HABER,  on  reducing  alumina 658 

Hall,  C.  M.,  on  reducing  alumina 658 

Halske  and  Siemens'  copper  process 555 

Heat  balance  sheet  of  Bessemer  converter 353 

Heat  balance  sheet  of  blast  furnace 281 

Heat  balance  sheet  of  electrical  furnace 

Heat  consumed,  in  reductions 292 

Heat  of  formation  of  chemical  compounds 18-40 

Heat  of  formation  of  slag. 500 

Heat  of  formation  of  some  nitrates 510 


668  INDEX 

Heat  of  formation  of  some  oxides 579 

Heat  of  formation  of  some  sulfides 578 

Heat  properties  of  matte 500 

Heat  properties  of  slag 500 

Heat  units 14 

Heats  of  formation ' •  !.-...  18 

Herreshoff  furnace,  calculations  on 489,  515 

Hess,  H.  D.,  on  open-hearth  furnaces 395 

Hixon's  notes  on  matte  smelting , .  .  526 

Hoepfner's  copper  chloride  process 557,  558 

Hofman's  Metallurgy  of  Lead 593 

Holla  way,  J.,  on  bessemerizing  matte .  525 

Holla  way,  J.,  on  oxidation  of  sphalerite 621 

Hot  blast,  sensible  heat  in 285 

Howe,  H.  M.,  notes  on  a  bessemer  charge 336 

Hydrates,  heats  of  formation  of 19 

Hydrides,  heats  of  formation  of 24 

Hydrocarbons,  combustion  of 45,  217 

Hydrocarbons,  heats  of  formation  of 24,  217 

Hydrogen  compounds,  heats  of  formation  of 18-40 

Hydrogen  compounds,  thermophysics  of 131-144 

Hydrogen,  thermophysics  of 75 

Hyposulfites,  thermophysics  of 136 

ICE,  specific  heat  of 119 

Ideal  working  of  a  blast  furnace,  Gruner's 274 

Ingalls,  on  cost  of  treating  zinc  ores 621 

Ingalls,  on  oxidation  of  sphalerite 621 

Insoluble  anodes,  use  of 551 

Iodides,  thermophysics  of 133 

Iodine,  thermophysics  of 95 

lola,  Kansas,  natural  gas  from 636 

Iridium,  thermophysics  of 99 

Iron  alloys,  thermophysics  of 113 

Iron  compounds,  heats  of  formation  of 18-40 

Iron  compounds,  thermophysics  of 125, 128-130 

Iron,  electric  refining  of 446 

Iron,  heat  of  oxidation  of 352,  366 

Iron  ores,  electrical  reduction  of 429 

Iron  oxide,  discussion  of  reduction  of 69 

Iron  oxide,  problems  concerning  reduction  of 128, 129 

Iron  oxides,  heat  absorbed  in  reducing '.  292 

Iron  persulfate  process  for  copper  ores 555 

Iron  sulf ate  solutions,  resistivity  of 559 

Iron  sulfide,  heat  absorbed  in  reducing 293 

Iron,  thermophysics  of 85 

Iron,  use  as  soluble  anode 548 

Isabella,  Tenn.,  pyritic  smelting  at •  489 

Isabella,  Tenn.,  smelting  of  matte  at. 515 

JANNETTAZ,  "Les  Cenvertisseurs  pour  le  Cuivre" 526 

Joplin,  zinc  ore  from 620 

Jiiptner,  data  on  furnace  charges 403 

KOCH,  on  warm  blast  in  smelting •  •  •  \  ^7 

Krafft,  on  the  boiling  of  silver  and  gold .  614 

LABORATORY  of  open-hearth  furnaces 391 

Laboratory  of  furnaces,  efficiency  of .' . .  414 

La  Lustre  smelter,  warm  blast  at 487 


INDEX  669 

Landis,  W.  S.,  laboratory  determinations  by 500 

Latent  heats  of  fusion,  discussion  of 71 

Latent  heats  of  fusion,  tables  of 75-103, 118 

Latent  heats  of  vaporization,  discussion  of 

Latent  heats  of  vaporization,  tables  of 75-103, 118-144 

Lead  alloys,  thermophysics  of 109 

Lead  compounds,  heats  of  formation  of 18-40 

Lead  compounds,  thermophysics  of 131-144 

Lead,  electrometallurgy  of 598 

Lead,  metallurgy  of ,          578 

Lead,  physical  constants  of '         583 

Lead,  refining  of 583 

Lead,  thermophysics  of 101 

Lead,  use  as  insoluble  anode 551 

Lead,  vapor  tension  of 

Lime,  amount  needed  in  Bessemer  converter 347 

Lime,  heat  in  preheated 356 

Lithium  compounds,  heats  of  formation  of 18-40 

Lithium  compounds,  thermophysics  of 131-144 

Lithium,  thermophysics 76 

MAGNESIUM  compounds,  heats  of  formation  of 18-40 

Magnesium  compounds,  thermophysics  of 131-144 

Magnesium,  thermophysics  of 80 

Magnus,  B.,  on  energy  losses  at  electrical  contacts 562 

Manganese,  behavior  in  the  blast  furnace 240 

Manganese  compounds,  heats  of  formation  of 18-40 

Manganese  compounds,  thermophysics  of 131-144 

Manganese,  heat  of  oxidation  of 371 

Manganese,  oxides,  heat  absorbed  in  reducing 

Manganese,  thermophysics  of 85 

Manhes,  P.,  on  bessemerizing  matte 525 

Marchesi,  on  electrolytic  treatment  of  matte 539,  541 

Marsh  gas,  mean  specific  heat  of 142 

Martin  process 

Matte,  bessemerizing  of 525 

Matte,  blister  roasting  of 523 

Matte,  bringing  forward  of 514 

Matte,  copper,  composition  of 469 

Matte,  electric  furnace  production  of 574 

Matte,  electrolytic  treatment  of 539 

Matte,  grade  obtained  in  smelting 474 

Matte,  thermophysical  properties  of 500 

Maximum  tension  of  aqueous  vapor 

Mayer,  F.,  Das  Bessemern  von  Kupf  ersteinen  " 526 

Mercury  alloys,  thermophysics  of 113 

Mercury  compounds,  heats  of  formation  of 18-40 

Mercury  compounds,  thermophysics 131-144 

Mercury,  metallurgy  of 653 

Mercury,  thermophysics  of 100 

Mercury,  vapor  tension  of 584 

Metallic  mist 656 

Methane,  mean  specific  heat  of 

Mine  water,  extraction  of  copper  from 

Mist,  metallic 656 

Moisture,  heat  absorbed  in  decomposing 293 

Molybdenum  compounds,  heats  of  formation  of 18-40 

Molybdenum  compounds,  thermophysics  of 131-144 

Molybdenum,  thermophysics  of ; 91 

Mond  gas 169 


670  INDEX 

Mond  gas,  superheating  of 175 

Monell  process,  problem  concerning 424 

Morgan  producer,  problem  concerning 163 

Mount  Lyell,  gases  of  furnace  at 497 

Mount  Lyell,  pyritic  smelting  at 488 

NATURAL  gas,  combustion  of 10,  47,  53,  219 

Natural  gas,  from  lola,  Kansas 636 

Niagara  Falls,  Salem  process  as  used  at 599 

Nickel  alloys,  thermophysics  of 113 

Nickel,  behavior  in  the  blast  furnace :  240 

Nickel  compounds,  heats  of  formation  of 18-40 

Nickel  compounds,  thermophysics  of 131-144 

Nickel,  heat  of  oxidation  of 353,  374 

Nickel,  thermophysics  of 86 

Nickeliferous  pyrrhotite,  reduction  of 435 

Nitrates,  heats  of  formation 30,  609 

Nitrates,  thermophysics  of 137 

Nitrides,  heats  of  formation  of 23 

Nitrogen,  thermophysics  of 78 

Oil,  fuel,  combustion  of 219 

Oil  furnace,  efficiency  of 106 

Open-hearth  furnaces 381 

Ore,  behavior  of,  in  the  blast  furnace 239 

Ore,  comparison  of  values  of 265 

Ore,  electrical  reduction  of 429 

Ores,  copper,  direct  treatment  by  electrolysis 536 

Osmium,  thermophysics  of 98 

Oxides,  heat  absorbed  in  reducing 292 

Oxides,  heats  of  formation  of ' 18,  579 

Oxides,  thermophysics  of 118 

Oxygen,  thermophysics  of 79 

PALLADIUM   compounds,  heats  of  formation  of 18-40 

Palladium,  thermophysics  of 92 

Paraffin,  thermophysics  of 142 

Parrot  Works,  Butte,  Bessemer  process  at 525 

Peat,  kiln  drying  of 227 

Peters,  "Modern  Copper  Smelting" 469,  526 

Peters,  "Principles  of  Copper  Smelting" 489,  484,  566 

Phosphates,  heats  of  formation  of 33 

Phosphates,  thermophysics  of 138 

Phosphides,  heats  of  formation  of 

Phosphorus,  behavior  in  the  blast  furnace 240 

Phosphorus,  heat  of  oxidation  of 353,  377 

Phosphorus,  oxide,  heat  absorbed  in  reducing 293 

Phosphorus,  thermophysics  of 81 

Pig-iron,  heat  of  formation  of 285 

Pig-iron,  heat  in  melted 289 

Platinum  alloys,  thermophysics  of 113 

Platinum  compounds,  heats  of  formation  of. 18-40 

Platinum  compounds,  thermophysics  of 131-144 

Platinum,  thermophysics  of 99 

Ports  of  an  open-hearth  furnace 388 

Potassium  compounds,  heats  of  formation  of 18-40 

Potassium  compounds,  thermophysics  of . 131-144 

Potassium,   thermophysics  of 83 

Pot  roasting,  applicability  to  sphalerite : 621 

Pot  roasting,  principle  of 586 


INDEX 


671 


Potter,  H.  N.,  on  heat  ( 
Powdered  coal,  combust 
Power  obtainable  from  : 
Power  required  to  comp 
Power,  surplus,  from  bl; 
Pressure  of  blast,  metric 
Pressure  of  blast,  requii 

Page 

Problem    1  8 
Problem    2  10 
Problem    3  11 
Problem    4  47 
Problem    5  54 
Problem    6  106 
Problem    7  107 
Problem    8  108 
Problem    9  114 
Problem  10               114 

)f  formation  of  s 
ion  of 

ilica 

353 
,223 
502 
313 
272 
317 
340 
Page 
460 
461 
461 
462 
462 
464 
465 
475 
477 
479 
482 
489 
497 
501 
510 
515 
530 
537 
541 
546 
549 
550 
552 
556 
558 
560 
563 
568 
573 
586 
593 
599 
602 
606 
610 
621 
624 
630 
633 
635 
636 
647 
650 
651 
653 
660 

229 
381 

483 

221 

furnace  gases.  .  . 

iress  blast 

ast  furnace  gase; 
>ds  of  measuring 
•ed  in  the  Bessei 

Problem    51  . 

s  

ner  proces 
Page 
245 

s      

Problem  94  
Problem  95  
Problem  96  
Problem  97  
Problem  98  
Problem  99  
Problem  100  
Problem  101  
Problem  102 

Problem    52. 
Problem    53  . 
Problem    54  . 
Problem    55 

259 
....   267 
.  .  .  .  268 
270 

Problem    56  . 
Problem    57. 
Problem    58. 
Problem    59. 
Problem    60. 
Problem    61  . 
Problem    62. 
Problem    63  . 
Problem    64  . 
Problem    65. 
Problem    66  . 
Problem    67. 
Problem    68  . 
Problem    69 

.  .  .  .  -272 
....   294 
....   305 
....  315 
319 
.  322 
....  327 
....   331 
....  334 
336 
.  ...   343 
.  ...  349 
.  ...  363 
384 

Problem  103. 

Problem  11  
Problem  12  
Problem  13  
Problem  14  
Problem  15  
Problem  16  
Problem  17.  ... 
Problem  18  
Problem  19.  ... 
Problem  20.  ... 
Problem  21  
Problem  22  
Problem  23  
Problem  24 

.  .   114 
.  .    115 
..    127 
..    128 
.  .    129 
.  .    149 
.  .    154 
.  .   163 
.  .    169 
.  .   179 
.  .    182 
.  .    192 
.  .    194 
208 

Problem  104  
Problem  105  
Problem  106  
Problem  107 

Problem  108  
Problem  109  
Problem  110  
Problem  111  

Problem  112 

Problem    70  . 
Problem    71  . 
Problem    72  . 
Problem    73. 
Problem    74. 
Problem    75. 
Problem    76  . 
Problem    77  . 
Problem    78. 
Problem    79  . 
Problem    80  . 
Problem    81  . 
Problem    82  . 
Problem    83  . 
Problem    84  . 
Problem    85  . 
Problem    86  . 
Problem    87  . 
Problem    88  . 
Problem    89  . 
Problem    90  . 
Problem    91 

.  ...   388 
.   393 
.  ...   403 
.  ...  411 
.  ...   418 
424 
.  ...   430 
.  ...  435 
.  ...   442 
443 
.  ...  446 
.  ...   452 
.  ...   452 
.  ...   453 
.  ...   454 
.  ...   454 
.  ...   455 
.  ...   456 
.  ...  456 
.  ...   456 
.  ...   457 
457 

Problem  113  
Problem  114  
Problem  115  
Problem  116  
Problem  117  
Problem  118  
Problem  119  
Problem  120  
Problem  121 

Problem  25  
Problem  26  
Problem  27  
Problem  28  
Problem  29.  ... 
Problem  30  
Problem  31  
Problem  32  

.  .   215 
.  .   219 
.  .   219 
.  .   220 
.  .   221 
.  .    221 
.  .    222 
.  .   223 

Problem  122  
Problem  123  
Problem  124  

Problem  125  
Problem  126 

Problem  33  
Problem  34  
Problem  35  
Problem  36.  ... 
Problem  37  
Problem  38  
Problem  39  
Problem  40  
Problem  41  
Problem  42.  ... 
Problem  43  
Problem  44.  ... 

.  .   223 
.  .   223 
.  .   224 
.  .   224 
.  .   225 
.  .   225 
.  .   226 
.  .   226 
.  .   227 
.  .   227 
.  .   228 
.  .   229 

Problem  127  
Problem  128  
Problem  129  

Problem  130  
Problem  131  ..  . 

Problem  132  
Problem  133  
Problem  134 

Problem    92  . 
Problem    93  . 

tion  of  

458 
.  ...   460 

Problem  135  
Problem  136  
Problem  137  
Problem  138  
Problem  139 

Problem  45 

229 

Problem  46     . 

.     230 

Problem  47.  ... 

PRODUCER  gas, 
Producers  gas 

.  .   231 

combus 

.  .219,227,228, 

246 

Pvritic  smelting 

672  INDEX 

Pyritic  smelting,  reaction  at  the  focus 498 

Pyritic  smelting,  slag  produced  by , 487 

Pyritic  smelting,  temperature  attained  in 485 

Pyrrhotite,  electrical  reduction  of 435 

Pyrrhotite,  smelting  of 489 

QUENEAU,  on  furnace  efficiency 416,  423 

RADIATION,  heat 213 

Radiation,  from  a  Bessemer  converter 533 

Radiation,  from  a  blast  furnace 289 

Radiation,  from  an  electrolytic  furnace 660 

Radiation,  from  a  roasting  furnace 479,  482,  627 

Radiation,  from  a  smelting  furnace 506 

Radiation,  heat,  tables  of ; 213 

Radiation,  heating  by 392 

Radium,  thermophysics  of 102 

Reactions,  double,  in  copper  metallurgy .   .      474 

Reactions,  double,  in  lead  metallurgy 582 

Recarburizatipn,  in  the  Bessemer  converter 348 

Reduction  of  iron  ores  electrically 429 

Reduction  of  roasted  lead  ore 591 

Reduction  of  zinc  oxide 627 

Refining,  electrolytic,  of  copper 558 

Refining,  electrolytic,  of  gold 610 

Refining,  electrolytic,  of  iron 446 

Refining,  electrolytic,  of  lead 598 

Refining,  electrolytic,  of  silver 605 

Refining,  electrolytic,  principles  of 536 

Refining,  of  impure  lead 583 

Regenerators,  dimensions  of 382 

Regenerators,  efficiency  of 412 

Resistivity  of  some  solutions 559 

Reverberatory  smelting 501 

Rhodium,  thermophysics  of 92 

Richard,  T.  A.,  on  pyrite  smelting 

Roasting,  Bessemer,  of  lead  ores 586 

Roasting,  heat  generated  in 477,  479 

Roasting  of  lead  sulfide 581,  586 

Roasting  of  sphalerite 620 

Roasting,  removal  of  sulfur  in 473 

Roasting,  smelting  of  copper  matte 523 

Rubber,  thermophysics  of .  142 

Rubidium,  thermophysics  of 90 

Ruthenium,  thermophysics  of :  .  .  . 91 

SALOM,  P.  G.,  process  for  copper  matte ', 545 

Salom,  P.  G.,  process  for  reducing  galena 598 

Saulte  Ste.  Marie,  electrical  reduction  at , 435 

Savelsberg  pot-roasting  operation 586 

Schnabel  "Handbook  of  Metallurgy" 469,  526 

Schuller,  on  the  boiling  of  silver  and  gold .  615 

Selenides,  heats  of  formation  of 

Selenium,  thermophysics  of 89 

Siemens  and  Halske's  copper  process 555 

Siemens-Martin  process 381 

Siemens  new-style  furnace : 

Silica,  behavior  of,  in  the  blast  furnace 240 

Silicates,  heats  of  formation  of 31 


INDEX  673 

Silicates,  thermophysics  of 139 

Silicides,  heats  of  formation  of 26 

Silicon  compounds,  heats  of  formation  of 18-40 

Silicon  compounds,  thermophysics  of 139 

Silicon,  heat  of  oxidation  of 353,  369 

Silicon,  thermophysics  of 81 

Silver  alloys,  thermophysics  of : 113 

Silver,  analyses  of  impure 606 

Silver,  behavior  in  the  blast  furnace 240 

Silver,  compounds,  heats  of  formation  of 18-40 

Silver  compounds,  thermophysics  of 131-144 

Silver,  metallurgy  of : 605 

Silver,  thermophysics  of : 93 

Silver,  vapor  tensions  of 616 

Slag,  amount  produced  in  the  blast  furnace 251 

Slag,  blast  furnace,  composition  of 252 

Slag,  blast  furnace,  heat  in '•'. 289 

Slag  formed  in  the  Bessemer  converter 346 

Slag,  fusion  of,  in  the  copper  furnace 500 

Slag,  heat  of  formation  of 286,  357 

Slag,  summation  of  ingredients  of •        254 

Slag,  thermophysics  of 144 

Smelting  electric  of  copper  ores 572 

Smelting  of  lead  ores .:.........  591 

Smelting  of  zinc  ores ,  637 

Smelting,  ordinary,  of  copper  ores . 499 

Smelting,  pyritic,  rate  of 

Snyder,  F.  T.,  on  efficiency  of  electric  furnaces 640 

Snyder,  P.  T.,  on  electric  zinc  smelting 641 

Sodium  compounds,  heats  of  formation  of 18-40 

Sodium  compounds,  thermophysics  of 131-144 

Sodium,  thermophysics  of 79 

Solutions,  extraction  of  copper  from 548 

Specific  heats  of  chemical  compounds - 131-144 

Specific  heats  of  products  of  combustion 51 

Specific  heats  of  the  elements 75-103 

Speiss,  formation  of 592 

Spence  furnace,  calculations  on  a 479 

Sphalerite,  roasting  of 620 

Steel,  electrical  production  of 440 

Steel  furnace,  efficiency  of 

Steel,  thermophysics  of 113 

Sticht,  analysis  of  furnace  gases  by 497 

Sticht,  on  pyritic  smelting 484 

Stolberg,  Marchesi's  process  at ^ 539 

Strontium  compounds,  heats  of  formation  of 18-40 

Strontium  compounds,  thermophysics  of. 131-144 

'Strontium,  thermophysics  of '.  .  . 90 

Sulfates,  heats  of  formation  of 

Sulfates,  thermophysics  of 136 

Sulfides,  heats  of  formation  of 21,  578 

Sulfides,  thermophysics  of 134 

Sulfur,  available  in  ordinary  smelting 474 

Sulfur,  available  in  pyritic  smelting •. 486 

Sulfur,  behavior  in  blast  furnace 237 

Sulfur,  condition  of,  in  roasted  ore 475,  479,  507 

Sulfur,  heat  absorbed  in  separation  of 293 

Sulfur,  removal  by  partial  roasting 473 

Sulfur,  thermophysics  of 

Sulfuric  acid  solutions,  resistivity  of 559 

43 


674  INDEX 

TANTALUM,  thermophysics  of 97 

Tellurides,  heats  of  formation  of 22 

Tellurium  compounds,  heats  of  formation  of 18-40 

Tellurium,  thermophysics  of 96 

Temperature,  theoretical,  of  combustion 50 

Temperature,  thermochemistry  of  high 61 

Temperatures  in  the  Bessemer  converter 368 

Temperatures  in  the  blast  furnace 308 

Temperatures,  using  dry  blast 312 

Tension,  maximum,  of  aqueous  vapor. 121 

Thallium  compounds,  heats  of  formation  of 18-40 

Thallium  compounds,  thermophysics  of 131-144 

Thallium,  thermophysics  of 101 

Thermochemical  constants  of  bases  and  acids 38 

Thermochemistry,  applications  of . . . 13 

Thermochemistry,  calculations  in 41 

Thermochemistry  of  the  Bessemer  process 351 

Thermophysics  of  alloys 109-1 13 

Thermophysics  of  chemical  compounds 118-144 

Thermophysics  of  the  elements 75-103 

Thermit  process,  calculations 43 

Thermit  process,  temperatures  in 

Thorium,  thermophysics  of 103 

Tin  alloys,  thermophysics  of 109 

Tin  compounds,  heats  of  formation  of 18-40 

Tin  compounds,  thermophysics  of. 131-144 

Tin,  thermophysics  of 94 

Tissier,  reduction  of  lead  oxide  by  aluminium 585 

Titanates,  heats  of  formation  of 34 

Titanium,  heat  of  oxidation  of 353,  373 

Titanium,  thermophysics  of 

Toldt,  data  on  furnace  charge 403 

Tungstates,  heats  of  formation  of 

Tungsten  compounds,  thermophysics  of . . '. 131-144 

Tungsten,  heat  of  oxidation  of 353 

Tungsten,  thermophysics  of 97 

ULKE,  modern  electrolytic  copper  refining 469 

Units,  heat,  definition  of 14 

Uranium,  thermophysics  of 103 

VACUUM,  volatility  of  gold  and  silver  in  a 614 

Valves  for  open-hearth  furnaces 

Vanadium,  heat  of  oxidation  of 353 

Vanadium,  thermophysics  of 84 

Vancouver,  B.  C.,  electric  zinc  smelting  at 641 

Van  Liew,  on  bessemerizing  matte 530 

Vapor,  aqueous,  maximum  tension  of 121 

Vaporization,  latent  heats  of,  discussion 73 

Vaporization,  latent  heats  of,  tables 75-103, 118-144 

Vapor,  metallic,  calculation  of  weight  of 618 

Vapor,  tension  of  cadmium 644 

Vapor,  tension  of  gold 615 

Vapor,  tension  of  lead 583 

Vapor,  tension  of  mercury 

Vapor,  tension  of  silver 615 

Vapor,  tension  of  zinc 643 

Vattier,  on  electric  smelting  of  copper  ores 573 

Volatility  of  lead 583 

Volatility  of  silver  and  gold *     614 


INDEX  675 

Voltages  of  decomposition 547 

Vulcanite,  thermophysics  of 142 

WATER,  cooling,  heat  in 290 

Water  gas 177 

Water  gas,  combustion  of 219 

Water,  heat  of  formation  of 15 

Water,  thermophysics  of 119 

Water  vapor,  maximum  tension  of 121 

Watts,  O.  P.,  on  boilingpoints  of  silver  and  gold 615 

Weightman,  A.  T.,  on  treatment  of  copper  matte 545 

Welch  process  of  copper  concentration 514 

Wohlwill  process  of  refining  gold  bullion 610 

Work  done  by  blowing  engines 313 

ZINC  alloys,  thermophysics  of 110 

Zinc,  behavior  in  the  blast  furnace 240 

Zinc  compounds,  heats  of  formation  of 18-40 

Zinc  compounds,  thermophysics  of 131-144 

Zinc,  distillation  of 632 

Zinc,  distilling  furnace,  efficiency  of 108 

Zinc,  electric  reduction  of 637 

Zinc,  heat  of  oxidation  of .  353 

Zinc,  metallurgy  of 620 

Zinc  oxide,  blast-furnace  reduction  of 651 

Zinc  oxide,  electric  reduction  of 626 

Zinc  oxide,  reduction  of 63,  627 

Zinc  oxide,  specific  heat  of 621 

Zinc,  reduction  in  a  blast  furnace 651 

Zinc  sulfide,  roasting  of 620 

Zinc  sulfide,  specific  heat  of 621 

Zinc  sulfide,  temperature  of  roasting 621 

Zinc,  thermophysics  of 87 

Zinc,  vapor  of 643 

Zirconium,  thermophysics  of 90 


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